M.C.Q (1 Marks) - Page 4 - MATHS STD 12 Science Questions - Vidyadip

Questions · Page 4 of 4

M.C.Q (1 Marks)

Question 1511 Mark

Choose the correct answer from the given four options. 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 2 green balls and one blue ball is:

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

Answer

$\frac{3}{28}$

Solution:
Probability of drawing 2 green balls and one blue ball
$=\text{P}_\text{G}\cdot\text{P}_\text{G}\cdot\text{P}_\text{B}+\text{P}_\text{B}\cdot\text{P}_\text{G}\cdot\text{P}_\text{G}+\text{P}_\text{G}\cdot\text{P}_\text{B}\cdot\text{P}_\text{G}$
$=\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{2}{6}+\frac{2}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}+\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{2}{6}$
$=\frac{1}{28}+\frac{1}{28}+\frac{1}{28}=\frac{3}{28}$

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Question 1521 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 1531 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 1541 Mark

A fair die is tossed eight times. The probability that a third six is observed in the eight throw is:

$\frac{\text{ }^7\text{C}_2\times5^5}{6^7}$
$\frac{\text{ }^7\text{C}_2\times5^5}{6^8}$
$\frac{\text{ }^7\text{C}_2\times5^5}{6^6}$
$\text{None of these}$

Answer

$\frac{\text{ }^7\text{C}_2\times5^5}{6^8}$

Solution:
probability of getting $6=\text{p}=\frac{1}{6},\text{q}=\frac{5}{6}$
probability of getting third six in eight throw.
= probability of getting 2 sixes in first seven throw + probability of getting six in eight throw
$=\Big(\text{ }^7\text{C}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^5\Big)\big(\frac{1}{6}\big)$
$=\frac{\text{ }^7\text{C}_2\times5^5}{6^8}$

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Question 1551 Mark

A bag contain 4 white and 2 black balls. Two balls are drawn at random. The probability that they are of the same colour is ________.

$\frac{5}{7}$
$\frac{1}{7}$
$\frac{7}{15}$
$\frac{1}{15}$

Answer

$\frac{7}{15}$

Solution:
We assume that there are 4 white balls and 2 black balls.
There are $\big(\frac{6}{2}\big)=15$ total possible ways of drawing two balls from these given 6 balls.
We are interested in the event where the two drawn balls are of the same colour.
For this, we note that the number of ways of drawing 2 white balls is $\big(\frac{4}{2}\big)=6$ whereas the number of ways of drawing 2 black balls is$\big(\frac{2}{2}\big)=1.$
So, the probability that the two drawn balls are of the same colour is $\frac{6+1}{15}=\frac{7}{15}.$

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Question 1561 Mark

Choose the correct answer from the given four options.Three persons, A, B and C, fire at a target in turn, starting with A. Their probability
of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits
is:

0.024
0.188
0.336
0.452

Answer

0.188

Solution:
We have
$\text{P}(\text{A})=0.4,\text{P}(\bar{\text{A}})=0.6,\text{P}(\text{B})=0.3,\text{P}(\bar{\text{B}})=0.7$
$\text{P}(\text{C})=0.2$ and $\text{P}(\bar{\text{C}})=0.8$
$\therefore$ Probability of two hits $=\text{P}_{\text{A}}\cdot\text{P}_{\text{B}}\cdot\text{P}_{\bar{\text{C}}}+\text{P}_{\text{A}}\cdot\text{P}{_\bar{\text{B}}}\cdot\text{P}_{\text{C}}+\text{P}{_\bar{\text{A}}}\cdot\text{P}_{\text{B}}\cdot\text{P}_{\text{C}}$
$=0.4\times0.3\times0.8+0.4\times0.7\times0.2+0.6\times0.3\times0.2$
$=0.096+0.056+0.036=0.188$

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Question 1571 Mark

In a class, 40% of the students study math and science. 60% of the students study math. What is the probability of a student studying science given he/she is already studying math?

0.43
0.40
0.67
0.60

Answer

0.67

Solution:
P(M and S) = 0.40
P(M) = 0.60
$\text{P(S|M})=\frac{\text{P (M and S)}}{\text{P(S)}}=\frac{0.40}{0.60}=\frac23=0.67$

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Question 1581 Mark

Choose the correct answer from the given four options.
A box has 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective?

$\Big(\frac{9}{10}\Big)^5$
$\frac{1}{2}\Big(\frac{9}{10}\Big)^4$
$\frac{1}{2}\Big(\frac{9}{10}\Big)^5$
$\Big(\frac{9}{10}\Big)^5+\frac{1}{2}\Big(\frac{9}{10}\Big)^4$

Answer

$\Big(\frac{9}{10}\Big)^5+\frac{1}{2}\Big(\frac{9}{10}\Big)^4$

Solution:
We have, $\text{n}=5,\text{P}=\frac{10}{100}=\frac{1}{10}$ and $\text{q}=\frac{9}{10}$
$\text{r}<1\Rightarrow\text{r}=0,1 $
Also, $\text{P}(\text{X}=\text{r})={^\text{n}}\text{C}_\text{r}\text{P}^\text{r}\text{q}^{\text{n}-\text{r}}$
$\therefore\text{P}(\text{X}=\text{r})=\text{P}(\text{r}=0)+\text{P}(\text{r}=1)$
$={^5}\text{C}_0\Big(\frac{1}{10}\Big)^0\Big(\frac{9}{10}\Big)^5+{^5}\text{C}_1\Big(\frac{1}{10}\Big)^1\Big(\frac{9}{10}\Big)^4$
$=\Big(\frac{9}{10}\Big)^5+5\cdot\frac{1}{10}\cdot\Big(\frac{9}{10}\Big)^4$
$=\Big(\frac{9}{10}\Big)^5+\frac{1}{2}\Big(\frac{9}{10}\Big)^4$

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Question 1591 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 1601 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 1611 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 1621 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 1631 Mark

Mark the correct alternative in the following question:
The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons 4 are swimmers is:

$\text{ }^5\text{C}_4(0.7)^4(0.3)$
$\text{ }^5\text{C}_1(0.7)(0.3)^4$
$\text{ }^5\text{C}_4(0.7)(0.3)^4$
$(0.7)^4(0.3)$

Answer

$\text{ }^5\text{C}_4(0.7)^4(0.3)$

Solution:
Given that a person is not a swimmer $\Rightarrow\text{q}=0.3$
$\Rightarrow\text{p}=0.7$
$\text{n = 5, X = 4}$
$\text{P(X}=4)=\text{ }^5\text{C}_4\times0.7^{4}\times0.3$

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Question 1641 Mark

An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?

0, 1, 2
3, 5, 7
7, 7, 8
1, 5, 7

Answer

0, 1, 2

Solution:
The two balls selected can be represented as BB, BR, RB, RR, where B represents a black ball and R represents a red ball.
X represents the number of black balls.
$\therefore$ X(BB) = 2
X(BR) = 1
X(RB) = 1
X(RR) = 0
Therefore, the possible values of X are 0, 1 and 2.
Yes, X is a random variable.

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Question 1651 Mark

A special lottery is to be held to select a student who will live in the only deluxe room in a hostel. There are 100 Year-III, 150 Year-II and 200 Year-I students who applied. Each Year-III's name is placed in the lottery 3 times; each Year-II's name, 2 times and Year-I's name, 1 time. What is the probability that a Year-III's name will be chosen?

$\frac18$
$\frac28$
$\frac38$
$\frac12$

Answer

$\frac38$

Solution:
Total names in the lottery
= 3 × 100 + 2 × 150 + 200 = 800
Number of Year-III's names = 3 × 100 = 300
Required probability $=\frac{300}{800}=\frac38$

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Question 1661 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 1671 Mark

Choose the correct answer from the given four options.
Let A and B be two events such that $\text{P}(\text{A})=\frac{3}{8},\text{P}({\text{B}})=\frac{5}{8}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{4}.$Then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}\Big(\frac{\text{A'}}{\text{B}}\Big)$ is equal to:

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

Answer

$\frac{6}{25}$

Solution:
We have, $\text{P}(\text{A})=\frac{3}{8},\text{P}({\text{B}})=\frac{5}{8}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{4}$
Now $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{3}{8}+\frac{5}{8}-\frac{3}{4}=\frac{1}{4}$
$\because\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{1}{4}}{\frac{5}{8}}=\frac{2}{5}$
and $\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)=\frac{\text{P}(\text{A}'\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{5}{8}-\frac{1}{4}}{\frac{5}{8}}=\frac{3}{5}$
$\therefore\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)$
$=\frac{2}{5}\cdot\frac{3}{5}=\frac{6}{25}$

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Question 1681 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 1691 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 1701 Mark

Choose the correct answer from the given four options.
If $\text{P}(\text{A})=\frac{4}{5},$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ then $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$ is equal to:

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

Answer

$\frac{7}{8}$

Solution:
$\text{P}(\text{A})=\frac{4}{5},\ \text{P}(\text{A}\cap\text{B})=\frac{7}{10}$
$\therefore\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}$
$=\frac{\frac{7}{10}}{\frac{4}{5}}=\frac{7}{8}$

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Question 1711 Mark

A four-digit number is formed by using the digits 1, 2, 4, 8 and 9 without repitition. If one number is selected from those numbers, then what is the probability that it will be an odd number?

$\frac15$
$\frac25$
$\frac35$
$\frac45$

Answer

$\frac25$

Solution:
Total number of outcomes = 5 × 4 × 3 × 2 = 120
The number of favourable cases = 2(4 × 3 × 2) = 48 (i.e., odd numbers)
Therefore,
Required probability $\frac{48}{120}=\frac25$

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Question 1721 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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M.C.Q (1 Marks) - Page 4 - MATHS STD 12 Science Questions - Vidyadip