3 Marks Question

Question 513 Marks

Two dice are thrown. Find the probability that the numbers appeared has the sum 8, if it is known that the second die always exhibits 4.

Answer

Two dice are thrown.
A = Sum on the dice is 8
A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
B = Second die always exhibits 4
B = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)}
$(\text{A}\cap\text{B})=\{(4, 4)\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{1}{6}$
Required probability $=\frac{1}{6}$

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Question 523 Marks

If A and B be two events such that $\text{P(A)}=\frac{1}{4},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{2},$ show that A and B are independent events.

Answer

$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{4}+\frac{1}{3}-\frac{1}{2}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{3+4-6}{12}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{12}=\frac{1}{4}\times\frac{1}{3}=\text{P(A)}\text{ P(B)}$
Thus, A and B are independent events.

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Question 533 Marks

If A and B are events such that P(A) = 0.6, P(B) = 0.3 and $\text{P}(\text{A}\cap\text{B})=0.2$ find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big).$

Answer

Given:
P(A) = 0.6
P(B) = 0.3
$\text{P}(\text{A}\cap\text{B})=0.2$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0.2}{0.3}=\frac{2}{3}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$\Rightarrow\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{0.2}{0.6}=\frac{1}{3}$

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Question 543 Marks

A coin is tossed three times. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ in each of the following:
A = At least two heads,
B = At most two heads.

Answer

Sample space for three coins is given by {HHH, HTH, THH, TTH, HHT, HTT, THT, TTT} A = At least two heads A = {HHT, HHT, HTH, THH} B = At most two heads B = {HHT, HTT, THT, TTT, HTH, THH, TTH} $(\text{A}\cap\text{B})=\{\text{HHT, HTH, THH}\}$ $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$ $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{3}{7}$Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{3}{7}$

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Question 553 Marks

An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the sum of the numbers obtained is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered $2, 3, 4, ....., 12$ is picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8?

Answer

Let $E_1, E_2$ and A be the events as defined below:
$E_1 =$ The coin shows a head
$E_2 =$ The coin shows a head
$A =$ The noted number is 7 or 8
$\therefore\ \text{P}(\text{E})_1=\frac{1}{2}$
$\text{P}(\text{E})_2=\frac{1}{2}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{11}{36}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{11}$
Using the law of total probability, we get
Required probability $=\text{P(A)}=\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{1}{2}\times\frac{11}{36}+\frac{1}{2}\times\frac{2}{11}$
$=\frac{11}{72}+\frac{1}{11}$
$=\frac{121+72}{792}$
$=\frac{193}{792}$

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Question 563 Marks

A die is thrown three times. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$, if
A = 4 appears on the third toss,
B = 6 and 5 appear respectively on first two tosses.

Answer

Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and and 5 on second throw
Clearly,
A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4), (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4), ...... (6, 1, 4), (6, 2 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
Now,
$(\text{A}\cap\text{B})=\{(6, 5, 4)\}$
$\therefore\ \text{Required probability} = \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{6}$
$\therefore\ \text{Required probability} = \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{A})}=\frac{1}{36}$

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Question 573 Marks

A coin is tossed three times, if head occurs on first two tosses, find the probability of getting head on third toss.

Answer

Consider the given events.
A = Getting head on third toss
B = Getting head on first two tosses
Clearly,
A = {(H, H, H), (H, T, H), (T, H, H), (T, T, H)}
B = {(H, H, H), (H, H, T)}
Now,
$\text{A}\cap\text{B}=\{\text{H},\text{H},\text{H}\}$
$\therefore\text{Required probability}=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{2}$

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Question 583 Marks

There are 3 red and 5 black balls in bag 'A'; and 2 red and 3 black balls in bag 'B'. One ball is drawn from bag 'A' and two from bag 'B'. Find the probability that out of the 3 balls drawn one is red and 2 are black.

Answer

It si givem that bag A contains 3 red and 5 balck balls (3R, 5B) and bag B contains 2 red and 3 black balls (2R, 3B).
Now,
P(One red and 2 black) = P(one red from bag A and two black from bag B) + P(black ball from bag A and remaining balls from bag B)
$=\frac{3}{8}\times\frac{3}{5}\times\frac{2}{4}+\frac{5}{8}\times\frac{2}{5}\times\frac{3}{4}\times2$
$=\frac{9}{80}+\frac{30}{80}$
$=\frac{39}{80}$
Note: 2 is multiplied by second term because there are two ways to select red and black balls from bag B.
While the first way is to pick black ball first, followed by red, the second way is to pick red ball first, followed by black.

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Question 593 Marks

Three machines $E_1, E_2, E_3$ in a certain factory produce $50 \%, 25 \%$ and $25 \%$, respectively, of the total daily output of electric bulbs. It is known that $4 \%$ of the tubes produced one each of the machines $E _1$ and $E _2$ are defective, and that $5 \%$ of those produced on $E_3$ are defective. If one tube is picked up at random from a day's production, then calculate the probability that it is defective.

Answer

Let D be the event that the picked up tube is defective.
Let $A_1, A_2$ and $A_3$ be the events that the tube is produced on macjines $E_1, E_2$ and $E_3$ respectively.
$P(D) = P(A_1) P(D|A_1) + P(A_2) P(D|A_2) + P(A_3) P(D|A_3) .....(i)$
$\text{P}(\text{A}_1)=\frac{50}{100}=\frac{1}{2},\text{P}(\text{A}_2)=\frac{25}{100}=\frac{1}{4},\text{P}(\text{A}_3)=\frac{25}{100}=\frac{1}{4}$
$\text{P}(\text{D}|\text{A}_1)=\text{P}(\text{D}|\text{A}_2)=\frac{4}{100}=\frac{1}{25}$
$\text{P}(\text{D}|\text{A}_3)=\frac{5}{100}=\frac{1}{20}$
Putting these values in (i), we get
$\text{P(D)}=\frac{1}{2}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{20}$
$\text{P(D)}=\frac{17}{400}$

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Question 603 Marks

A bag contains 5 white, 7 red and 3 black balls. If three balls are drawn one by one without replacement, find the probability that none is red.

Answer

Bag contains 5 white, 7 red and 3 black balls.
Total number of balls = 15
Three balls are drawn without replacement
A = first ball is red
B = Second ball is red
C = Third balls is red
P (Three balls are drawn, non is red)
$=\text{P}(\overline{\text{A}})\text{P}\Big(\overline{\frac{\text{B}}{\text{A}}}\Big)\text{P}\Big(\frac{\overline{\text{C}}}{\text{A}\cap\text{B}}\Big)$
$=\frac{8}{15}\times\frac{7}{14}\times\frac{6}{13}$
[Since, number of non red balls = 5 + 3 = 8]
$=\frac{8}{65}$
Required probability $=\frac{8}{65}$

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Question 613 Marks

Three cards are drawn with replacement from a well shuffled pack of 52 cards. Find the probability that the cards are a king, a queen and a jack.

Answer

Three cards are drawn with replacement fron a pack of cards
There are 4 Kings, 4 Queens, 5 Jacks.
P (1 King, 1 Queen, 1 Jack)
$=\text{P}\big[(\text{K}\cap\text{Q}\cap\text{J})\cup(\text{K}\cap\text{J}\cap\text{Q})\cup(\text{J}\cap\text{K}\cap\text{Q})\\\cup(\text{J}\cap\text{Q}\cap\text{K})\cup(\text{Q}\cap\text{K}\cap\text{L})\cup(\text{Q}\cap\text{J}\cap\text{K})\big]$
$=\text{P}(\text{K}\cap\text{Q}\cap\text{J})+\text{P}(\text{K}\cap\text{J}\cap\text{Q})+\text{P}(\text{J}\cap\text{K}\cap\text{Q})\\+\text{P}(\text{J}\cap\text{Q}\cap\text{K})+\text{P}(\text{Q}\cap\text{K}\cap\text{L})+\text{P}(\text{Q}\cap\text{J}\cap\text{K})$
$=\text{P}(\text{K}) \text{P}(\text{Q}) \text{P}(\text{J})+\text{P}(\text{K}) \text{P}(\text{J}) \text{P}(\text{Q})+\text{P}(\text{J}) \text{P}(\text{K}) \text{P}(\text{Q}) \\ +\text{P}(\text{J}) \text{P}(\text{Q}) \text{P}(\text{K})+\text{P}(\text{Q}) \text{P}(\text{K}) \text{P}(\text{L})+\text{P}(\text{Q}) \text{P}(\text{J}) (\text{K})$
$=\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52} \\ +\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}$
$=\frac{6}{13\times13\times13}$
$=\frac{6}{2197}$
Required probability $=\frac{6}{2197}$

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MATHS 3 Marks Question