M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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50 questions · 7 auto-graded MCQ + 43 self-marked written.

Question 11 Mark

Out of 30 consecutive integers, 2 are chosen at random. The probability that their sum is odd, is

$\frac{14}{29}$
$\frac{16}{29}$
$\frac{15}{29}$
$\frac{10}{29}$

Answer

$\frac{15}{29}$

Solution:
For sum of two integers to be odd, one integer should be even and the other should be odd. In 30 consecutive integers, 15 are even and 15 are odd.
P(Sum is odd) = P(first integer is odd and second is even) + P(first integer is even and second integer is odd)
$=\frac{15}{30}\times\frac{15}{29}+\frac{15}{30}\times\frac{15}{29}$
$=\frac{450}{30\times29}$
$=\frac{15}{29}$

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Question 21 Mark

If A and B are two events such that $\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{4}.$ and $\text{P}(\text{A}|\text{B})\times\text{P}(\overline{\text{A}}\cap\text{B})$ is equals to.

$\frac{2}{5}$
$\frac{3}{8}$
$\frac{3}{20}$
$\frac{6}{25}$

Answer

$\frac{6}{25}$

Solution:
$\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{8},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{3}{4}$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \frac{3}{4}=\frac{3}{8}+\frac{5}{8}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}\times\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}\times\frac{\text{P(B)}-\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{5}{8}}\times\frac{\big(\frac{5}{8}-\frac{1}{4}\big)}{\frac{5}{8}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{6}{25}$

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Question 31 Mark

A bag contains 5 brown and 4 white socks. A man pulls out two socks. The probability that these are of the sane colour is.

$\frac{5}{108}$
$\frac{18}{108}$
$\frac{30}{108}$
$\frac{48}{108}$

Answer

$\frac{48}{108}$

Solution:
Total number of balls = 5brown + 4white = 9
Required probability $=\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times\frac{3}{8}=\frac{4}{9}$
$\Rightarrow\ \frac{4\times12}{9\times12}=\frac{48}{108}$

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Question 41 Mark

An urn contains 9 balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of the same colour is,

$\frac{5}{84}$
$\frac{3}{9}$
$\frac{3}{7}$
$\frac{7}{17}$

Answer

$\frac{5}{84}$

Solution:
Given:
Red balls = 2
Blue balls = 3
Black balls = 4
P(All three balls are of same colour) = P(all three are blue) + P(all three are black)
$=\frac{3}{9}\times\frac{2}{8}\times\frac{1}{7}+\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7}$
$=\frac{1}{84}+\frac{4}{84}$
$=\frac{5}{84}$

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MCQ 51 Mark

A die is thrown and a card is selected ar random from a deck pf $52$ playing cards. The probability of getting an even number of the die and a spade card is

A
$\frac{1}{2}$
B
$\frac{1}{4}$
✓
$\frac{1}{8}$
D
$\frac{3}{4}$

Answer

Correct option: C.

$\frac{1}{8}$

A Sample space when a die is thrown,
$S_1 = \{1, 2, 3, 4, 5, 6\} $
$\Rightarrow n(S_1) = 6$
Let $A$ be the event that getting even number.
$A = \{2, 4, 6\} $
$\Rightarrow n(A) = 3$
$\Rightarrow\ \text{P(A)}=\frac{3}{6}=\frac{1}{2}$
$A$ card is selected from a deck of $52$ cards.
$\text{n}(\text{S}_2)= {^{52}}\text{C}_2=52$
Let $B$ be the event that getting spade card.
$\text{n(B)}= {^{13}}\text{C}_2=13$
$\Rightarrow\ \text{P(B)}=\frac{13}{52}=\frac{1}{4}$
Required probability $= P(A) \times P(B)$
$=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}$

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MCQ 61 Mark

Five persone entered the lift cabin on the ground floor of an $8$ floor house. Suppose that each of them independently and with equal probability can leave the cabin at any flor beginning with the first, then the probability of all $5$ persons leaving at different floors is,

✓
$\frac{^{7}\text{P}_5}{7_5}$
B
$\frac{7_5}{^{7}\text{P}_5}$
C
$\frac{6}{^{6}\text{P}_5}$
D
$\frac{^{5}\text{P}_5}{5}$

Answer

Correct option: A.

$\frac{^{7}\text{P}_5}{7_5}$

Five persons can leave different floors
By $^7P_5$ ways.
Possible ways of leavinf the lift $= 7^5$
Required probability $=\frac{^{7}\text{P}_5}{7^5}$

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Question 71 Mark

If two events are independent, then.

They must be mutually exclusive.
The sum of their probabilities must be equal to 1.
(a) and (b) both are correct.
None of the above is correctIf two. events are independent, then.

Answer

None of the above is correctIf two. events are independent, then.

Solution:
Let A and B are two independent events, Then,
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}$
As, $\text{P}(\text{A}\cap\text{B})\neq0\text{ or }\text{P(A)}+\text{P(B)}\neq1$
So, both are neither mutually exclisive nor their sum of probability is 1.
Hence, the correct alternative is option (d).

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Question 81 Mark

In a college 30% students fail in Physics, 25% fail in Mathenatics and 10% fail in both. One student is chosen at random. The probability that she fails in Physics if she failed in Mathematics is.

$\frac{1}{10}$
$\frac{1}{3}$
$\frac{2}{5}$
$\frac{9}{20}$

Answer

$\frac{2}{5}$

Solution:
Let A be the event that students failed in Physics. B be the event that students failed in Mathematics.
Given that, $\text{P(A)}=30\%=\frac{30}{100}$
$\text{P(B)}=25\%=\frac{25}{100}$
$\text{P}(\text{A}\cap\text{B})=10\%=\frac{10}{100}$
Required probability is given by $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{\frac{10}{100}}{\frac{25}{100}}=\frac{2}{5}$

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Question 91 Mark

If $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\overline{\text{A}\cap\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})=$

$\frac{1}{5}$
$\frac{4}{5}$
$\frac{1}{2}$
$1$

Answer

1

Solution:
$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{4}{5}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P}(\overline{\text{A}\cup\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})$
$=1-\text{P}(\text{A}\cap\text{B})+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=1-\frac{3}{10}+\frac{3}{5}-\frac{3}{10}$
$=1$

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MCQ 101 Mark

Two persons $A$ and $B$ take turns in throwing a pair of dice.The first person to throw $9$ from both dice will be awarded the prize. If $A$ throws first, then the probability that $B$ wins the game is.

A
$\frac{9}{17}$
✓
$\frac{8}{17}$
C
$\frac{8}{9}$
D
$\frac{1}{9}$

Answer

Correct option: B.

$\frac{8}{17}$

$9$ can be obtained from throw of two dice in only $4$ cases as given below:
$\{(3, 6), (4, 5), (5, 4), (6, 3)\}$
$\Rightarrow\ \text{P(getting }9)=\frac{4}{36}=\frac{1}{9}$
$\text{P(not getting }9)=\frac{32}{36}=\frac{8}{9}$
Now,
$P(B$ is winning$) = P($getting $9$ in $2^{nd}$ throw$) + P($getting $p$ in $4^{th}$ throw$) + P($getting $9$ in $6^{th}$ throw$) + .....$
$=\frac{8}{9}\times\frac{1}{9}+\frac{8}{9}\times\frac{8}{9}\times\frac{8}{9}\times\frac{1}{9}+\ .....$
$=\frac{8}{81}\Big[1+\frac{64}{81}+\Big(\frac{64}{81}\Big)^2+\ ......\Big]$
$=\frac{8}{81}\times\frac{1}{1-\frac{64}{81}}$
$=\frac{8}{81}\times\frac{81}{17}$
$=\frac{8}{17}$

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Question 111 Mark

If $\text{P}(\text{A}\cup\text{B})=0.8$ and $\text{P}(\text{A}\cap\text{B})=0.3$ then $\text{P}(\overline{\text{A}})=\text{P}(\overline{\text{B}})=$

0.3
0.5
0.7
0.9

Answer

0.9

Solution:
If $\text{P}(\text{A}\cup\text{B})=0.8\text{ P}(\text{A}\cap\text{B})=0.3,$
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=\text{P}(\text{A}\cup\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=1.1$
$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$
$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=\big[\text{P(A)}+\text{P(B)}\big]$
$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-1.1$
$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=0.9$

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MCQ 121 Mark

A bag $X$ contains $2$ white and $3$ black balls and another bag $Y$ contains $4$ white and $2$ black balls. One bag is selected at random and a ball is drawn from it. Then, the probability chosen to be white is,

A
$\frac{2}{15}$
B
$\frac{7}{15}$
C
$\frac{8}{15}$
✓
$\frac{14}{15}$

Answer

Correct option: D.

$\frac{14}{15}$

A white ball can be drawn in two mutually exclusive ways:

Selecting bag $X$ and then drawing a white ball from it.
Selecting bag $Y$ ane then drawing a white ball from it.

Let $E_1, E_2$ and $A$ be the three evenes as defined below:
$E_1 =$ Selecting abg $X$
$E_2 =$ Selecting bag $Y$
$A =$ Drawing a white ball
We know that one bag is selected randomly.

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Question 131 Mark

Let A and B be two events such that P(A) = 0.6, P(B) = 0.2, P(A|B) = 0.5. Then $\text{P}(\overline{\text{A}}|\overline{\text{B}})$ equals.

$\frac{1}{10}$
$\frac{3}{10}$
$\frac{3}{8}$
$\frac{6}{7}$

Answer

$\frac{3}{8}$

Solution:
Given that,
$\text{P(A)}=0.6,\text{P(B)}=0.2,\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.5$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.5$
$\Rightarrow\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=0.5$
$\Rightarrow\frac{\text{P}(\text{A}\cap\text{B})}{0.2}=0.5$
$\Rightarrow \text{P}(\text{A}\cap\text{B})=0.1$
$\Rightarrow \text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=0.1$
$\Rightarrow \text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-0.1$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=0.7$
Now, $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$
$\Rightarrow \text{P}(\overline{\text{A}}\cap\overline{\text{B}})=0.3$
To find
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{\text{P}(\overline{\text{A}\cap\text{B}})}{\text{P}(\overline{\text{B}})}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{0.3}{0.8}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{3}{8}$

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Question 141 Mark

Three persons, A, B and C fine a target in turn starting with A. Their probability of hitting the target are 0.4, 0.2 and 0.2, respectively. The probability of two hits is

0.024
0.452
0.336
0.188

Answer

0.188

Solution:
Let:
A be the event of hitting the target by the person A,
B be the event of hitting the target by the person B and
C be the event of hitting the target by the person C
We have,
P(A) = 0.4, P(B) = 0.3 and P(C) = 0.2
Also,
$\text{P}(\overline{\text{A}})=1-\text{P(A)}=1-0.4=-0.6,$
$\text{P}(\overline{\text{B}})=1-0.3=0.7$ and
$\text{P}(\overline{\text{C}})=1-0.2=0.8$
Now,
$\text{P(Two hits)}=\text{P}(\text{AB}\overline{\text{C}})+\text{P}(\text{A}\overline{\text{B}}\text{C})+\text{P}(\overline{\text{A}}\text{BC})$
$=\text{P(A)}\times\text{P(B)}\times\text{P}(\overline{\text{C}})+\text{P(A)}\times(\overline{\text{B}})\times\text{P(C)}\\+\text{P}(\overline{\text{A}})\times\text{P(B)}\times\text{P(C)}$
$=0.4\times0.3\times0.8+0.4\times0.7\times0.2+0.6\times0.3\times0.2$
$=0.096+0.056+0.036$
$=0.188$
Hence, the correct alternative is option (d).

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Question 151 Mark

If A and B are two events, then $\text{P}(\overline{\text{A}}\cap\text{B})=$

$\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})$
$1-\text{P}(\text{A})-\text{P}(\text{B})$
$\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

Answer

$\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

Solution:

From the diagram, we get $\text{A}\cap\text{B}$ and $\overline{\text{A}}\cap\text{B}$ are mutually exclusive events such that $(\text{A}\cap\text{B})\cup(\overline{\text{A}}\cap\text{B})=\text{B}.$ therefore by
$\text{P}(\text{A}\cap\text{B})+\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}$
$\therefore\ \text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$

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MCQ 161 Mark

The probability that in a year of $22^{nd}$ century chosen at random, there will be $53$ Sunday, is

A
$\frac{3}{28}$
B
$\frac{2}{28}$
✓
$\frac{7}{28}$
D
$\frac{5}{28}$

Answer

Correct option: C.

$\frac{7}{28}$

We know a leap year is fallen within $4$ years,
So its probability is $\frac{25}{100}=\frac{1}{4}$
$53^{rd}$ Sunday leap year $=\frac{1}{4}\times\frac{2}{7}=\frac{2}{28}$
Similarly probability of $53^{rd}$ Sunday in a non leap year
$=\frac{75}{100}\times\frac{1}{7}=\frac{3}{4}\times\frac{1}{7}=\frac{3}{28}$
Required probability $=\frac{2}{28}+\frac{3}{28}=\frac{5}{28}$.

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Question 171 Mark

India play two matches each with West indies and Australia. In any match the probability of india getting 0,1 and 2 points are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are indepecdent, the probability of india getting at least 7 points is.

0.0875
$\frac{1}{16}$
0.1125
None of these.

Answer

0.0875

Solution:
Here, there are total 5 ways by which India can get at least 7 points.

2 points + 2 points + 2 points + 2 points = (0.5 × 0.5 × 0.5 × 0.5)
1 points + 2 points + 2 points + 2 points = (0.05 × 0.5 × 0.5 × 0.5)
2 points + 1 points + 2 points + 2 points = (0.5 × 0.05 × 0.5 × 0.5)
2 points + 2 points + 1 points + 2 points = (0.5 × 0.5 × 0.05 × 0.5)
2 points + 2 points + 2 points + 1 points = (0.5 × 0.5 × 0.5 × 0.05)

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Question 181 Mark

A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 22 green balls and one blue ball is

$\frac{167}{168}$
$\frac{1}{28}$
$\frac{2}{21}$
$\frac{3}{28}$

Answer

$\frac{3}{28}$

Solution:
Total balls in a box - 3orange + 3green + 2blue = 8
Three balls are drawn at random from the box then samplw space $\text{n(S)}= {^{8}}\text{C}_3=\frac{8\times7\times6}{3\times2\times1}=56$
Let A be the event that drawing 2 green and one blue ball.
$\text{n(A)}={^{3}}\text{C}_2\times{^{2}}\text{C}_2=6$
$\text{P(A)}=\frac{6}{56}=\frac{3}{28}$

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Question 191 Mark

A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one iten is chosen ar random, the probability that it is rusted or is nail is

$\frac{3}{16}$
$\frac{5}{16}$
$\frac{11}{16}$
$\frac{14}{16}$

Answer

$\frac{11}{16}$

Solution:
Rusted items = 3 + 5 = 8
Rusted nails = 3
Total nails = 6
P(getting a rusted item or a nail) = P(getting a rusted item) + P(getting a nail) - P(getting a rusted item and a nail)
$=\frac{8}{16}+\frac{6}{16}-\frac{3}{16}$
$=\frac{8+6-3}{16}$
$=\frac{11}{16}$

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Question 201 Mark

If the events A and B are independent, then $\text{P}(\text{A}\cap\text{B})$ is equal to,

P(A) + P(B)
P(A) - P(B)
P(A) P(B)
$\frac{\text{P(A)}}{\text{P(B)}}$

Answer

P(A) P(B)

Solution:
$\text{P}(\text{A}\cap\text{B})=\text{P(A)} \text{ P(B)}$ for independent events.

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Question 211 Mark

If A and B are two events associated to a random experiment such that $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$ and $\text{P(B)}=\frac{17}{20}$, then P(A|B) =

$\frac{14}{17}$
$\frac{17}{20}$
$\frac{7}{8}$
$\frac{1}{8}$

Answer

$\frac{14}{17}$

Solution:
$\text{P}(\text{A}\cap\text{B})=-\frac{7}{10},\text{P(B)}=\frac{17}{20}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{7}{10}}{\frac{17}{20}}=\frac{14}{17}$

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Question 221 Mark

If A and B are two events such that $\text{P(A)}\neq0$ and $\text{P(B)}\neq1,$ then $\text{P}(\overline{\text{A}}|\overline{\text{B}})=$

$1-\text{P}(\text{A}|\text{B})$
$1-\text{P}(\overline{\text{A}}|\text{B})$
$\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{A}})}{\text{P}(\overline{\text{B}})}$

Answer

$\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$

Solution:
We have,
$\text{P(A)}\neq0$ and $\text{P(B)}\neq1$
Now,
$\text{P}(\overline{\text{A}}|\overline{\text{B}})=\frac{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}{\text{P}(\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{A}\cap\text{B}})}{\text{P}(\overline{\text{B}})}$
$=\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
Hence, the correct alternative is option (C).

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Question 231 Mark

If A and B are two events such that $\text{P}(\text{A}|\text{B})=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9},$ then p =

$\frac{2}{3}$
$\frac{3}{5}$
$\frac{1}{3}$
$\frac{3}{4}$

Answer

$\frac{1}{3}$

Solution:
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3},\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p}$
$\Rightarrow \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\text{P}$
$\Rightarrow \frac{\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})}{\text{P(B)}}=\text{P}$
$\Rightarrow\frac{\text{p}+\frac{1}{3}-\frac{5}{9}}{\frac{1}{3}}=\text{p}$
$\Rightarrow\text{p}+\frac{1}{3}-\frac{5}{9}=\frac{\text{p}}{3}$
$\Rightarrow\frac{-2}{9}=\frac{\text{p}}{3}-\text{p}$
$=\frac{-2}{3}\text{p}=\frac{-2}{9}$
$\Rightarrow\text{p}=\frac{1}{3}$

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Question 241 Mark

A flash light has 8 batteries out of which 3 are dead. IF two batteries are selected without replacement and tested, then the probability that both are dead is,

$\frac{3}{28}$
$\frac{1}{14}$
$\frac{9}{64}$
$\frac{33}{56}$

Answer

$\frac{3}{28}$

Solution:
We have,
The total number of batteries = 8
The number of dead batteries = 3
Let A be the event of selecting the first dead battery and B be the event of selecting the second dead battery.
Now,
P(both dead batteries are selected) $=\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}\times\text{P}(\text{B}|\text{A})$
$=\frac{3}{8}\times\frac{2}{7}$
$=\frac{3}{28}$
Hence, the correct alternative is option (a).

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Question 251 Mark

Associated to a random experiment two events A and B are such that $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$. The value pf P(A) is

$\frac{3}{10}$
$\frac{1}{2}$
$\frac{1}{10}$
$\frac{3}{5}$

Answer

$\frac{1}{2}$

Solution:
$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{3}{10}$
$\text{P(A)}+\frac{3}{5}-\frac{4}{5}=\frac{3}{10}$
$\text{P(A)}=\frac{1}{2}$

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Question 261 Mark

A box contains 10 good articles and 6 with defects. One item is drawn at random. The probability that it is either good or has a defect is,

$\frac{64}{64}$
$\frac{49}{64}$
$\frac{40}{64}$
$\frac{24}{64}$

Answer

$\frac{64}{64}$

Solution:
P(good item) $=\frac{10}{16}$
P(defected item) $=\frac{6}{16}$
P(eitherr good or defected item) = P(good item) + P(defected item)
$=\frac{10}{16}+\frac{6}{16}$
$=\frac{16}{16}$
$=1$
$=\frac{64}{64}$

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Question 271 Mark

If A and B are two events such that $\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3},\text{P}(\text{A}|\text{B})=\frac{1}{4},$ then $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$ equals.

$\frac{1}{12}$
$\frac{3}{4}$
$\frac{1}{4}$
$\frac{3}{16}$

Answer

$\frac{1}{4}$

Solution:
$\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{4}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{1}{3}}=\frac{1}{4}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{12}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{12}}{\frac{1}{3}}=\frac{1}{4}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=-1\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\Big[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\Big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{1}{4}$

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Question 281 Mark

A and B draw two cards each, one after another, from a pack of well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is

$\frac{44}{85\times49}$
$\frac{11}{85\times49}$
$\frac{13\times24}{17\times25\times49}$
None of these.

Answer

$\frac{44}{85\times49}$

Solution:
Total cards = 52 There are four suits of cards in a pack, i.e. diamond, heart, spade and club.
Pall 4 cards are of same suit = Pall 4 cards are of diamond + Pall 4 cards are of heart + Pall 4 cards are of spade + Pall 4 cards are of club.
$=4\times\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{59}$
$=4\times\frac{11}{85\times49}$
$=\frac{44}{85\times49}$

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Question 291 Mark

If A and B are two events such that $\text{P(A)}=\frac{4}{5},$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ then P(B|A) =

$\frac{1}{10}$
$\frac{1}{8}$
$\frac{7}{8}$
$\frac{17}{20}$

Answer

$\frac{7}{8}$

Solution:
We have,
$\text{P(A)}=\frac{4}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$
Now,
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\Big(\frac{7}{10}\Big)}{\Big(\frac{4}{5}\Big)}$
$=\frac{7\times5}{10\times4}$
$=\frac{7}{8}$
Hence, the correct alternative is option (c).

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Question 301 Mark

If A and B are two independent events such that P(A) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5,$ then P(A|B) - P(B|A) =

$\frac{2}{7}$
$\frac{3}{35}$
$\frac{1}{70}$
$\frac{1}{7}$

Answer

$\frac{1}{70}$

Solution:
We have,
$\text{P(A)}=0.3$ and $\text{P}(\text{A}\cup\text{B})=0.5$
As, A and B are independent events
So, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}$
$=0.3\times\text{P(B)}$
$=0.3\text{ P(B)}\ .....\text{(i)}$
Also, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow 0.5 = 0.3+\text{P(B)}-0.3\text{ P(B)}$ [Using (i)]
$\Rightarrow 0.5-0.3 = 0.7\text{ P(B)}$
$\Rightarrow0.7\text{ P(B)}=0.2$
$\Rightarrow\text{ P(B)}=\frac{0.2}{0.7}$
$\Rightarrow\text{ P(B)}=\frac{2}{7}$
Using (i), we get
$\text{P}(\text{A}\cap\text{B})=0.3\times\frac{2}{7}=\frac{6}{70}$
Now,
$\text{P}(\text{A}|\text{B})-\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}-\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\Big(\frac{6}{70}\Big)}{\Big(\frac{2}{7}\Big)}-\frac{\Big(\frac{6}{70}\Big)}{0.3}$
$=\frac{6\times7}{70\times2}-\frac{6}{70\times0.3}$
$=\frac{3}{10}-\frac{2}{7}$
$=\frac{21-20}{70}$
$=\frac{1}{70}$
Hence, the correct alternative is option (c).

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Question 311 Mark

If A and B are such that $\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$ and $\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3},$ then $\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=$

$\frac{9}{10}$
$\frac{10}{9}$
$\frac{8}{9}$
$\frac{9}{8}$

Answer

$\frac{10}{9}$

Solution:
$\text{P}(\text{A}\cup\text{B})=\frac{5}{9},\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
Consider,
$\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\Rightarrow\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
$\Rightarrow 1-\text{P}(\text{A}\cap\text{B})=\frac{2}{3}$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}-\frac{5}{9}=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=\frac{8}{9}$
$\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\big[\text{P(A)}+\text{P(B)}\big]$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\frac{8}{9}$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=\frac{10}{9}$

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Question 321 Mark

If one ball is drawn ar random from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, then the probability that 2 white and 1 black balls will be drawn is.

$\frac{13}{32}$
$\frac{1}{4}$
$\frac{1}{32}$
$\frac{3}{16}$

Answer

$\frac{13}{32}$

Solution:
Total balls in first box = 3 white + 1 black = 4
Total balls in second box = 2 white + 2black = 4
Total balls in third box = 1white + 3black = 4
Probability of 2 white and 1 black
= P(WWB) + P(WBW) + P(BWW)
$=\frac{3}{4}\times\frac{2}{4}\times\frac{3}{4}+\frac{3}{4}\times\frac{2}{4}\times\frac{1}{4}+\frac{1}{4}\times\frac{2}{4}\times\frac{1}{4}$
$=\frac{18+6+2}{64}=\frac{13}{32}$

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Question 331 Mark

From a set of 100 cards numbered 1 to 100, one card is drawn at randow. The probability number obtained on the card is divisible by 6 or 8 but not by 24 is

$\frac{6}{25}$
$\frac{1}{4}$
$\frac{1}{6}$
$\frac{2}{6}$

Answer

$\frac{6}{25}$

Solution:
Number of cards divisible by 6 = 16
$\Rightarrow\ \text{P(A)}=\frac{16}{100}$
Number of cards divisible by 8 = 12
$\Rightarrow\ \text{P(B)}=\frac{12}{100}$
Number of cards divisible by 24 = 4
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{4}{100}$
$\text{P}(\text{A}\cup\text{B})=\frac{16}{100}+\frac{12}{100}-\frac{4}{100}$
$\text{P}(\text{A}\cup\text{B})=\frac{6}{25}$

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Question 341 Mark

A bag containe 5 black, 4white balls and 3 red balls. if a ball is selected randomwise, the probability that it is black or red ball is,

$\frac{1}{3}$
$\frac{1}{4}$
$\frac{5}{12}$
$\frac{2}{3}$

Answer

$\frac{2}{3}$

Solution:
We know that the bag contains 5B (black), 4W(white) and 3R(red) balls.
Now,
$\text{P(B)}=\frac{5}{12}$
$\text{P(R)}=\frac{3}{12}$
$\text{P}(\text{B or R})=\text{P(B)}+\text{P(R)}$
$=\frac{5}{12}+\frac{3}{12}$
$=\frac{8}{12}=\frac{2}{3}$

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Question 351 Mark

A coin is tossed three times. If events A and B are defined as A = Two heads come, B = Last should be head, Then, A and B are

Independent.
Dependent.
Both.
Mutually exclusive.

Answer

Dependent.

Solution:
S = [(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)]
$\text{P(A)}=\text{P}(2\text{heads})=\frac{3}{8}$
$\text{P(B)}=\text{P}(\text{last one is heads})=\frac{4}{8}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{8}=\frac{1}{4}\neq\text{P(A) P(B)}$
Thus, A and B are dependent.

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Question 361 Mark

A and B are two events such that P(A) = 0.25 and P(B) = 0.50. The probability pf both happening together is 0.14. The probability of both A and B hot happening is.

0.39
0.25
0.11
None of these.

Answer

0.39

Solution:
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.25+0.5-0.14$
$0.61$
P(Both A and B not happening) $=\text{P}(\text{A}\cup\text{B})'$
$=1-\text{P}(\text{A}\cup\text{B})$
$=1-0.61$
$=0.39$

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Question 371 Mark

If A and B are two independent events with $\text{P(A)}=\frac{3}{5}$ and $\text{P(B)}=\frac{4}{9},$ then $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$ equals,

$\frac{4}{15}$
$\frac{8}{45}$
$\frac{1}{3}$
$\frac{2}{9}$

Answer

$\frac{2}{9}$

Solution:
$\text{P(A)}=\frac{3}{5},\text{P(B)}=\frac{4}{9}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\Big[\frac{3}{5}+\frac{4}{9}-\frac{3}{5}\times\frac{4}{9}\Big]$
($\because$ A and B are independent)
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\frac{7}{9}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{2}{9}$

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Question 381 Mark

The probability that a leap year will have 53 fridays or 53 Saturdays is.

$\frac{2}{7}$
$\frac{3}{7}$
$\frac{4}{7}$
$\frac{1}{7}$

Answer

$\frac{3}{7}$

Soluction:
Non-leap year has 365 days = 52 weeks + 1
366 days in leap year.
We want to find probability of 53 Fridays or 53 Saturday.
Favourable cases = {(Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
Required probability $=\frac{3}{7}$

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MCQ 391 Mark

Two cards are drawn from a well shuffled deck of $52$ playing cards with replacement. The probability that both cards are queen is

✓
$\frac{1}{13}\times\frac{1}{13}$
B
$\frac{1}{13}+\frac{1}{13}$
C
$\frac{1}{13}\times\frac{1}{17}$
D
$\frac{1}{13}\times\frac{4}{5}$

Answer

Correct option: A.

$\frac{1}{13}\times\frac{1}{13}$

Two cards are drawn from $52$ cards.
Let, $E_1$ be the event that getting queen in first draw and $E_2$ be the event that getting queen in second draw,
$\text{P}(\text{E}_1\cap\text{E}_2)=\frac{4}{52}\times\frac{4}{52}=\frac{1}{13}\times\frac{1}{13}$

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Question 401 Mark

If $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\text{B}|\overline{\text{A}})=$

$\frac{1}{5}$
$\frac{3}{10}$
$\frac{1}{2}$
$\frac{1}{2}$

Answer

$\frac{3}{5}$

Solution:
$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{3}{10}$
$\text{P(A)}+\frac{3}{5}-\frac{4}{5}=\frac{3}{10}$
$\text{P(A)}=\frac{1}{2}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{\text{P}(\text{B}\cap\overline{\text{A}})}{\text{P}(\overline{\text{A}})}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{\frac{3}{5}-\frac{3}{10}}{1-\frac{1}{2}}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{3}{5}$

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Question 411 Mark

If S is the samle space and $\text{P(A)}=\frac{1}{3}, \text{P(B)}$ and $\text{S}=\text{A}\cup\text{B,}$ where A and B are tow mutually exclusive events, then P(A) =

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$
$\frac{3}{8}$

Answer

$\frac{1}{4}$

Solution:
$\text{P(A)}=\frac{1}{3}\text{P(A)}$
$\Rightarrow\ \text{P(B)}=3\text{P(A)}\ .....(\text{i})$
A and B are mutually exclusive events.
$\Rightarrow\ \text{P}(\text{A}\cap\text{B}) = 0$
Now,
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}=\text{P(S)}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=1$
$\Rightarrow\ \text{P(A)}+3\text{P(A)}=1$ [From (i)]
$\Rightarrow\ 4\text{P(A)}=1$
$\Rightarrow\ \text{P(A)}=\frac{1}{4}$

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MCQ 421 Mark

Assume that in a family, each chold is equally likely to be a boy or a girl. A family with tree cgildren is chosen at random. Tere probability that the eldest child is a girl given that the family has at least oe girl.

A
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{2}{3}$
✓
$\frac{4}{7}$

Answer

Correct option: D.

$\frac{4}{7}$

$S = \{GBB, GGB, GBG, GGG, BGG, BGB, BBG, BBB\}$
Let $E_1$ be the event that choosing a family with a girl as eldest child.
$E_2$ be the event that choosing a family with at least one girl.
$E_1 = \{GBB, GGB, GBG, GGG\}$
$E_2 = \{GBB, GGB, GBG, GGG, BGG, BGB, BBG\}$
$\text{n}(\text{E}_1)=4,\text{n}(\text{E}_2)=7,\text{n}(\text{A}\cap\text{B})=4$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n(B)}}=\frac{4}{7}$

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Question 431 Mark

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, then the probability that exactly two of the three balls were red, the first ball being red, is

$\frac{1}{3}$
$\frac{4}{7}$
$\frac{15}{28}$
$\frac{5}{28}$

Answer

$\frac{4}{7}$

Solution:
Total number of balls = 5 red + 3 Blue = 8
Probability of getting exacctly two red balls given that first ball should be red
Required probability $=\text{P}\Big(\frac{\text{R}_2\text{B}_2}{\text{R}_1}\Big)+\text{P}\Big(\frac{\text{R}_1\text{B}_2}{\text{R}_1}\Big)$
Required probability $=\frac{4}{7}\times\frac{3}{6}+\frac{3}{7}\times\frac{4}{6}=\frac{4}{7}$

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Question 441 Mark

If P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6 then $\text{P}(\text{A}\cup\text{B})=$

0.24
0.3
0.48
0.96

Answer

0.96

Solution:
We have,
P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6
As, P(B|A) = 0.6
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}=0.6$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.6\times\text{P(A)}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.6\times0.4$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.24$
Now, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.4+0.8-0.24$
$=1.2-0.24$
$=0.96$
Hence, the correct alternative is option (d).

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Question 451 Mark

Three faces of aj ordinary dice are yellow, two faces are red and one face is blue. The dice is rolled 3 times. The probability that yellow red and blue face appear in the first second and third throws respectively, is

$\frac{1}{36}$
$\frac{1}{6}$
$\frac{1}{30}$
None of these.

Answer

$\frac{1}{36}$

Solution:
P(yellow face) $=\frac{3}{6}=\frac{1}{2}$
P(red face) $=\frac{2}{6}=\frac{1}{3}$
P(one face) $=\frac{1}{6}$
P(yellow face, red face and blue face appear in the required order) $=\frac{1}{2}\times\frac{1}{3}\times\frac{1}{6}=\frac{1}{36}$

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Question 461 Mark

A speaks truth in 75% cases and B seaks truth in 80% cases. Probability that they contradict each other in a statement, is

$\frac{7}{20}$
$\frac{13}{20}$
$\frac{3}{5}$
$\frac{2}{5}$

Answer

$\frac{7}{20}$

Soluction:
P(A speaks truth) = 0.75
P(A lies) = 1 - 0.75 = 0.25
P(B speaks truth) = 0.8
P(B lies) = 1 - 0.8 = 0.2
P(contradicting each other in a statement) = P(A speaks truth and lies) + P(B speaks truth and A lies)
= 0.75 × 0.2 + 0.8 × 0.25
= 0.15 + 0.2
= 0.35
$=\frac{35}{100}=\frac{7}{20}$

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Question 471 Mark

Two dice are thrown simultaneously. The probability of getting a pair of aces is

$\frac{1}{36}$
$\frac{1}{3}$
$\frac{1}{6}$
None of these.

Answer

$\frac{1}{36}$

Solution:
Required probability = Probability of ace in first throw + Probability of ace in second throw
$=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$

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Question 481 Mark

Let A and B be two events. If $\text{P(A)}=0.2,\text{P(B)}=0.4,\text{P}(\text{A}\cup\text{B})=0.6$ then P(A|B) is equal to

0.8
0.5
0.3
0

Answer

0

Solution:
$\text{P(A)}=0.2,\text{P(B)}=0.4,\text{P}(\text{A}\cup\text{B})=0.6$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{A}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0.2+0.4-0.6}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0$

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Question 491 Mark

If A and B are two events such that $\text{A}\neq\phi,\text{B}=\phi,$ then,

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{P(A)}\text{ P(B)}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=1$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P(A)}}{\text{P(B)}}$

Answer

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$

Solution:
If A and B are two events such that $\text{A}\neq\phi, \text{B}=\phi$ then,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$

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Question 501 Mark

The probabilities of a student getting I, II and III division in an examination are $\frac{1}{10},\frac{3}{5}$ and $\frac{1}{4}$ respectively. The probability that the student fails in the examination is.

$\frac{197}{200}$
$\frac{27}{100}$
$\frac{83}{100}$
None of these.

Answer

$\frac{27}{100}$

Solution:
$\text{P(A)}=\frac{1}{10},\text{P(B)}=\frac{3}{5},\text{P(C)}=\frac{1}{4}$
Required probability $=\text{P}(\overline{\text{A}}\cap\overline{\text{B}}\cap\overline{\text{C}})$
Required probability $=\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})\text{ P}(\overline{\text{C}})$
Required probability $=(1-\text{P(A)})(1-\text{P(B)})(1-\text{P(C)})$
Required probability $=\Big(1-\frac{1}{10}\Big)\Big(1-\frac{3}{5}\Big)\Big(1-\frac{1}{4}\Big)$
Required probability $=\frac{27}{100}$

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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip