2 Marks Questions

Question 512 Marks

A and B are two events such that P (A) ≠ 0. Find P(B|A), if

A is a subset of B
$\text{A}\cap\text{B}=\phi$

Answer

A and B are two events such that P (A) ≠ 0.
To find: P(B|A)

$\text{A is a subset of B}\ \Rightarrow\ \text{A}\subset\text{B}$

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

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

$\text{A}\cap\text{B}=\phi$

$\therefore\ \text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{0}{\text{P}(\text{A})}=0$

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Question 522 Marks

The probability distribution function oif a random variable X is given by

$X_i$	0	1	2
$P_i$	$3c^3$	$4c - 10c^2$	$5c - 1$

Where $c > 0$
Find: $P(X < 2)$.

Answer

$\text{P}(\text{X}<2)=\text{P}(0)+\text{P}(1)$
$=3\text{c}^3+4\text{c}-10\text{c}^2$
$=3\Big(\frac{1}{3}\Big)^3+4\Big(\frac{1}{3}\Big)-10\Big(\frac{1}{3}\Big)^2$
$=\frac{3}{27}+\frac{4}{3}-\frac{10}{9}$
$=\frac{1}{9}+\frac{4}{3}-\frac{10}{9}$
$=\frac{3}{9}$
$\therefore\ \text{P}(\text{x}<2)=\frac{1}{3}$

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Question 532 Marks

Determine P(E|F) in Exercises. A die is thrown three times,E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.

Answer

Since a dice has six faces. Therefore n (S) = 6 × 6 × 6 = 216
E = (1, 2, 3, 4, 5, 6) × (1, 2, 3, 4, 5, 6) × (4)
F = (6) × (5) × (1, 2, 3, 4, 5, 6) $\Rightarrow$ n(F) = 1 × 1 × 6 = 6
$\text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{6}{216}$
$\therefore\ \ \ \ \ \ \ \text{E}\cap\text{F}=\left(6, 5, 4\right)\ \ \ \ \ \Rightarrow\ \ \ \text{n}\left(\text{E}\cap\text{F}\right)=1$
$\therefore\ \ \ \ \ \ \ \text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\cap\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{1}{216}$
$\text{And}\ \ \ \ \text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\cap\text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{216}$

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Question 542 Marks

A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Of the same colour.

Answer

Bag contains 3 blue, 5 red marble. One marble is drawn, its colour noted and replaced then again a marble drawn and its colour is noted.
P (Of the same colour)
$=\text{P}\big((\text{R}_1\cap\text{R}_2)\cup(\text{B}_1\cap\text{B}_2)\big)$
$=\text{P}(\text{R}_1)\times\text{P}(\text{R}_2)+\text{P}(\text{B}_1)\times\text{P}(\text{B}_2)$
$=\frac{5}{8}\times\frac{5}{8}+\frac{3}{8}\times\frac{3}{8}$
$=\frac{25+9}{64}$
$=\frac{34}{64}$
$=\frac{17}{32}$

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Question 552 Marks

An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
One red and one blue ball.

Answer

An urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 blue balls)
$=\text{P}\big((\text{R}\cap\text{B})\cup(\text{B}\cap\text{R})\big)$
$=\text{P(R)}\times\text{P(B)}+\text{P(B)}\times\text{P(R)}$
$=\frac{7}{11}\times\frac{4}{11}+\frac{4}{11}\times\frac{7}{11}$
$=\frac{28+28}{121}$
$=\frac{56}{121}$
Required probability $=\frac{56}{121}$

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Question 562 Marks

If $\text{P(A)}=0.3,\text{P(B)}=0.6,\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5,$ find $\text{P}(\text{A}\cup\text{B}).$

Answer

Given,
$\text{P(A)}=0.3,\text{P(B)}=0.6,\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5,$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$0.5=\frac{\text{P}(\text{A}\cap\text{B})}{0.3}$
$\text{P}(\text{A}\cap\text{B})=0.5\times0.3$
$\text{P}(\text{A}\cap\text{B})=0.15$
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.3+0.6+0.15$
$=0.75$
$\text{P}(\text{A}\cup\text{B})=0.75$

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Question 572 Marks

The mean and variance of a binomial distribution are $\frac{4}{3}$ and $\frac{8}{9}$ respectively. Find P (X ≥ 1).

Answer

$=1-\big(\frac{2}{3}\big)^{4}$
$=1-\frac{16}{81}$
$=\frac{65}{81}$
$\text{P(X}\geq1)=\frac{65}{81}$

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Question 582 Marks

Three digit numbers are formed with the digits 0, 2, 4, 6 and 8. Write the probability of forming a three digit number with the same digits.

Answer

Total 3-digit numbers that can be made out of 0, 2, 4, 5 and 8 = 4 × 5 × 5 (hundreds place cannot be filled with 0)
= 100
But 222, 444, 666 and 888 are four numbers, which have the same digits at all places.
P(3-digit number having same digits at all places) $=\frac{4}{100}$
$=\frac{1}{25}$

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Question 592 Marks

An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
2 Blue balls.

Answer

An urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 blue balls)
$=\text{P}(\text{B}_1\cap\text{B}_2)$
$=\text{P}(\text{B}_1)\times\text{P}(\text{B}_2)$
$=\frac{4}{11}\times\frac{4}{11}$
$=\frac{16}{121}$
Required probability $=\frac{16}{121}$

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Question 602 Marks

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 ?

Answer

There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.
Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.

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Question 612 Marks

Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\overline{\text{A}}\cap\text{B})$

Answer

Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.6 - 0.18$
$\text{P}(\overline{\text{A}}\cap\text{B})=0.42$

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Question 622 Marks

A bag contains $4$ white balls and $2$ black balls. Another contains $3$ white balls and $5$ black balls. If one ball is drawn from each bag, find the probability that,
One is while and one is black.

Answer

Bag A has 4 white balls and 2 black balls;
Bag B has 3 white balls and 5 black balls;
$P (A_W$ and $B_B$ or $A_B$ and $B_W)$
$= P(A_W) P(B_B) + P(A_B) P(B_W)$
$=\frac{4}{6}\times\frac{5}{8}+\frac{2}{6}\times\frac{3}{8}$
$=\frac{20}{48}+\frac{6}{48}$
$=\frac{26}{48}=\frac{13}{24}$

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Question 632 Marks

If the probability distribution of a random variable X is given below:

$X = X_i$	1	2	3	4
$P(X = X_i)$	c	2c	4c	4c

Write the value of $\text{P}(\text{X}\leq2)$

Answer

We know that the sum of probabilities in a probability distribution is always 1.
Therefore,
$P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1$
$\Rightarrow c + 2c + 4c + 4c =1$
$\Rightarrow 11c = 1$
$\Rightarrow\text{c}=\frac{1}{11}$
Now,
$\text{P}(\text{X}\leq2)=\text{P}(\text{X}=1)+\text{P}(\text{X})=2=\frac{1}{10}+\frac{2}{10}=\frac{3}{10}$

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Question 642 Marks

A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, find the probability that these two balls are of the same colour.

Answer

Given:
Bag 1 = (3W + 6B) balls
Bag 2 = (5B + 4W) balls
P(balls of same colour are drawn) = P(both black) + P(both white)
$=\frac{6}{9}\times\frac{5}{9}+\frac{3}{9}\times\frac{4}{9}$
$=\frac{30}{81}+\frac{12}{81}$
$=\frac{42}{81}$
$=\frac{14}{27}$

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Question 652 Marks

A fair coin is tossed 8 times, find the probability of.
at least six heads.

Answer

Probability of getting atleast 6 heads
$=\text{P}(\text{X}\geq6)$
$=\text{P}(\text{X}=6)+\text{P}(\text{X}=7)+\text{P}(\text{X}=8)$
$=\text{ }^8\text{C}_6\big(\frac{1}{2}\big)^8+\text{ }^8\text{C}_7\big(\frac{1}{2}\big)^8+\text{ }^8\text{C}_8\big(\frac{1}{2}\big)^8$
$=(28+8+1)\times\frac{1}{256}$
$=\frac{37}{256}$

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Question 662 Marks

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

Answer

$\text{P (king)}=\frac{4}{52}$
$\text{P (queen)}=\frac{4}{52}$
$\text{P (jack)}=\frac{4}{52}$
These cards can be drawn in $^3P_3$ ways.
$\text{P (king, queen, and jack)}=\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}\times {^{3}}\text{P}_3$
$=\frac{3!}{2197}$
$=\frac{6}{2197}$

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Question 672 Marks

Six coins are tossed simultaneously. Find the probability of getting.
no heads.

Answer

Let p represents the probability of getting head in a toss of fair coin, so
$\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$ [Since p + q = 1]
$\text{q}=\frac{1}{2}$
Let X denote the random variable representing the number heads in 6 tosses of coin. probability of getting r sixes in n tosses of a fair coin is given by,
$\text{P(X = r})=\text{ }^\text{n}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$3
$=\text{ }^6\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}}\dots(1)$
P(getting no head) = P(X = 0)
$=\frac{\text{ }^6\text{C}_0}{2^6}$
$=\big(\frac{1}{2}\big)^6$
$=\frac{1}{64}$

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Question 682 Marks

If in a binomial distribution mean is 5 and variance is 4, write the number of trials.

Answer

$\text{Mean}=5$ and $\text{Variance}=4$
$\Rightarrow\text{np}=5$ and $\text{npq}=4$
$\Rightarrow\text{q}=0.8$
$\Rightarrow\text{p}=1-\text{q}=0.2$
$\&\text{ np}=\text{n}(0.2)=5$ (given)
$\Rightarrow\text{n}=\frac{5}{0.2}=25$

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Question 692 Marks

A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Write the probability that the number is divisible by 5.

Answer

To be divisibel by 5 ones place sholud be 5
There are 3 place remaining which can be filled in 3! = 6 ways
So, 6 number can be formed out of 1, 2, 3 and 5, which are divisible by 5.
Total 4 - digit numbers = 4! = 24
P (4-digit number divisible by 5) $=\frac{6}{24}$
$=\frac{1}{4}$

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Question 702 Marks

Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that
all the five cards are spades?

Answer

Let P denote the probability of geting one spade out of a deck of 52 cards, So
$\text{p}=\frac{13}{52}$
$\text{p}=\frac{1}{4}$
$\text{q}=1-\frac{1}{4}$ [Since p + q = 1]
$\text{q}=\frac{3}{4}$
let X denote the radom variable of number of spades out of 5 cards. probability of getting r spades out of n cards is given by
$\text{P}(\text{x = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^5\text{c}_\text{r}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{5-\text{r}}\dots(1)$
Probability of getting all five spads
$\text{P}(\text{X}=5)$
$=\text{ }^5\text{c}_5\big(\frac{1}{4}\big)^5\big(\frac{3}{4}\big)^5-5$
$=\frac{1}{1024}$
Probability of getting 5 spades $=\frac{1}{1024}$

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Question 712 Marks

A bag contains $4$ white balls and $2$ black balls. Another contains $3$ white balls and $5$ black balls. If one ball is drawn from each bag, find the probability that,
Both are white.

Answer

Bag A has 4 white balls and 2 black balls;
Bag B has 3 white balls and 5 black balls;
$P (A_W$ and $B_W) = P(A_W) P(B_W)$
$=\frac{4}{6}\times\frac{3}{8}=\frac{1}{4}$

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Question 722 Marks

Which of the following distributions of a random variable X are the probability distributions?

X:	0	1	2
P(X):	0.6	0.4	0.2

Answer

P(X = 0) + P(X = 1) + P(X = 1) + P(X = 2)
= 0.6 + 0.4 + 0.2
= 1.2 > 1
It is not the probability distribution of random variable X.

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Question 732 Marks

Which of the following distributions of a random variable X are the probability distributions?

X:	0	1	2	3	4
P(X):	0.1	0.5	0.2	0.1	0.1

Answer

P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 0.1 + 0.5 + 0.2 + 0.1 + 0.1
= 1
It is the probability distribution of random variable X.

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Question 742 Marks

Two cards are drawn from a well shuffled pack of 52 cards, one after another without replacement. Find the probability that one of these is red card and the other a black card?

Answer

P(One red and one black) = P(first red and second black) + P(first black and second red)
$=\frac{26}{5}\times\frac{26}{51}+\frac{26}{52}\times\frac{26}{51}$
[Without replacement]
$=\frac{13}{51}+\frac{13}{51}$
$=\frac{26}{51}$

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Question 752 Marks

A bag contains 2 white, 3 red and 4 blue balls. Two balls are drawn at random from the bag. If X denotes the number of white balls among the two balls drawn, describe the probability distribution of X.

Answer

$\text{X}$	$\text{P(X)}$
$0$	$\frac{7}{6}\times\frac{6}{8}=\frac{21}{36}$
$1$	$\frac{7}{9}\times\frac{2}{8}\times2=\frac{14}{36}$
$2$	$\frac{2}{9}\times\frac{1}{8}=\frac{1}{36}$

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Question 762 Marks

If on an average 9 ships out of 10 arrive safely at ports, find the mean and S.D. of the ships returning safely out of a total of 500 ships.

Answer

Total number of ships (n) = 500
Let X denote the number of ships returning safely to the ports.
$\text{p}=\frac{9}{10}$ and $\text{q}=1-\text{p}=\frac{1}{10}$
$\text{Mean = np = 450}$ and $\text{Variance = npq = 45}$
$\text{Mean = 450}$
$\text{Standard deviation}=\sqrt{45}=6.71$

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Question 772 Marks

Six coins are tossed simultaneously. Find the probability of getting.
at least one head.

Answer

Let p represents the probability of getting head in a toss of fair coin, so
$\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$ [Since p + q = 1]
$\text{q}=\frac{1}{2}$
Let X denote the random variable representing the number heads in 6 tosses of coin. probability of getting r sixes in n tosses of a fair coin is given by,
$\text{P(X = r})=\text{ }^\text{n}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$3
$=\text{ }^6\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}}\dots(1)$
P(getting at least 1 head) $=\text{P(X}\geq1)$
$=1-\text{P(X}=0)$
$=1-\frac{1}{64}$
$=\frac{63}{64}$

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Question 782 Marks

A card is drawn from a well-shulffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled. Another card is then drawn from the deck.
What is the probability that both the cards are of the same suit?

Answer

P(both the cards are of same suit) = P(both the cards are of diamond) + P(both the cards are of spade) + P(both the cards are of club) + P(both the cards are of hear)
$=\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}$
$=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}$
$=\frac{4}{16}$
$=\frac{1}{4}$

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Question 792 Marks

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
none.

Answer

Let X be the number of bulbs fuse after 150 days.
X follows a binomial disribution with n = 5, p = 0.05 and q = 0.95
Or $\text{p}=\frac{1}{20}$ and $\text{q}=\frac{19}{20}$
$\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{20}\big)^{\text{r}}\big(\frac{19}{20}\big)^{5-\text{r}}$
Probability (nine will fuse after 150 days of use) = P(X = 0)
$=\text{ }^5\text{C}_0\big(\frac{1}{20}\big)^0\big(\frac{19}{20}\big)^{5-0}$
$=\big(\frac{19}{20}\big)^5$

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Question 802 Marks

If $\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13},$ find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big).$

Answer

Given:
$\text{P(A)}=\frac{7}{13}$
$\text{P(B)}=\frac{9}{13}$
$\text{P}(\text{A}\cap\text{B})=\frac{4}{13}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{4}{13}}{\frac{9}{13}}=\frac{4}{9}$

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Question 812 Marks

Find the expected number of boys in a family with 8 children, assuming the sex distribution to be equally probable.

Answer

Here, $\text{n}=8$
Let p be the probability of number of boys in the family.
$\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$
Expected number of boys = Mean
$\Rightarrow\text{np}=4$

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Question 822 Marks

In a competition A, B and C are participating. The probability that A wins is twice that of B, the probability that B wins is twice that of C. Find the probability that A losses.

Answer

Let P(A wins) = x
So, P(B wins) = 2x
P(A wins) = 2P(B wins)
= 2(2x)
P(A wins) = 4X
P(A wins) + P(B wins) + P(C wins) = 1
⇒ 4x + 2x+ x = 1
⇒ 7x = 1
$\Rightarrow\ \text{x}=\frac{1}{7}$
$\text{P(A wins)}=4\text{x}$
$=\frac{4}{7}$
$\text{P(A losses)}=1-\text{P(A wins)}$
$=1-\frac{4}{7}$
$=\frac{3}{7}$
Required probability $=\frac{3}{7}$

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Question 832 Marks

A random variable has the following probability distribution:

$X = X_i$	1	2	3	4
$P(X = X_i)$	k	2k	3k	4k

Write the value of $\text{P}(\text{X}\geq3).$

Answer

Here,

$X = X_i$	1	2	3	4
$P(X = X_i)$	k	2k	3k	4k

Since, $\sum\text{P}(\text{X})=1$ ⇒ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1 ⇒ k + 2k + 3k + 4k =1 ⇒ 10k = 1 $\Rightarrow\text{k}=\frac{1}{10}$ $\text{P}(\text{X}\geq3)$= P(X = 3) + P(X = 4)
= 3k + 4k
= 7k $=\frac{1}{10}$ $\text{P}(\text{X}\geq3)=\frac{7}{10}$

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Question 842 Marks

A ordinary cube has four plane faces, one face marked $2$ and another face marked $3$, find the probability of getting a total of $7$ in $5$ throws.

Answer

A cube has total 6 faces.
Total possible outcomes in 5 throws = $6 \times 6\times 6 \times 6 \times 6 = (6)^5$
The only way of getting 7 is by getting two 2s and one 3.
Total possible ways $=\frac{\text{P}^5_3}{2!}$
$=\frac{5\times4\times3\times2\times1}{2\times1\times2\times1}$
$=30$
Now,
P(getting 7 in 5 throws) $=\frac{30}{6^5}=\frac{5}{6^4}$

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Question 852 Marks

Suppose that a radio tube inserted into a certain type of set has probability $0.2$ of functioning more than $500$ hours. If we test $4$ tubes at random what is the probability that exactly three of these tubes function for more than $500$ hours$?$

Answer

Let $X$ denote the number of tubes that function for more than $500$ hours.
Then, $X$ follows a binomial distribution with $n = 4.$
Let $p$ be the probability that the tubes function more than $500$ hours
Here, $p = 0.2, q = 0.8$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^4\text{C}_{\text{r}}(0.2)^{\text{r}}(0.8)^{4-\text{r}},\text{r}=0,1,2,3,4$
Therefore, required probability $= P(X = 3)$
$= 4(0.2)^3(0.8)$
$= 0.0256$

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Question 862 Marks

A random variable X has the following probability distribution:

Values of X:	-2	-1	0	1	2	3
P(X)	0.1	k	0.2	2k	0.3	k

Find the value of k.

Answer

Here,

Values of X:	-2	-1	0	1	2	3
P(X)	0.1	k	0.2	2k	0.3	k

We know that,
P(-2) + P(-1) + P(0) + P(1) + P(2) + P(3) = 1
⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1
⇒ 4k + 0.6 = 1
⇒ 4k = 1 - 0.6
⇒ 4k = 0.4
$\Rightarrow\text{k}=\frac{0.4}{4}$
$\Rightarrow\text{k}=\frac{1}{10}$
$\Rightarrow\text{k}=0.1$

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Question 872 Marks

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
First ball is black and second is red.

Answer

Given,
Box contains 10 black and 8 red balls.
Two balls are drawn with replacement.
P (First ball is black and second is red)
$=\text{P}(\text{B}\cap\text{R})$
$=\text{P(B)}\text{ P(R)}$
$=\frac{10}{18}\times\frac{8}{18}$
$=\frac{20}{81}$
Required probability $=\frac{20}{81}$

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Question 882 Marks

A factory produces bulbs. The probability that one bulb is defective is $\frac{1}{50}$ and they are packed in boxes of 10. From a single box, find the probability that.
more than 8 bulbs work properly.

Answer

Let getting a defective bulb from a single box is a success.
We have,
p = probability of getting a defective bulb $=\frac{1}{50}$
Also, $\text{q}=1-\text{p}=1-\frac{1}{50}=\frac{49}{50}$
Let X denote the number of success in a sample of 10 trials. Then,
X follows binomial distribution with parameters n = 10 and $\text{p}=\frac{1}{50}$
$\therefore\text{P(X = r})=\text{ }^{10}\text{C}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(10-\text{r})}=\text{ }^{10}\text{C}_{\text{r}}\big(\frac{1}{50}\big)^{\text{r}}\big(\frac{49}{50}\big)^{(10-\text{r})}=\frac{\text{ }^{10}\text{C}_{\text{r}}49^{(10-\text{r})}}{50^{10}},$ where r = 0, 1, 2, 3, .....,10
Now,
Required probility = P(more than 8 bulbs work properly)
= P(atmost one bulb is defective)
$=\text{P(X}\leq0)$
$=\text{P(X}=0)+\text{P(X}=1)$
$=\frac{\text{ }^{10}\text{C}_049^{10}}{50^{10}}+\frac{\text{ }^{10}\text{C}_149^9}{}$
$=\frac{49^{10}}{50^{10}}+\frac{10\times49^{9}}{50^{10}}$
$=\frac{49^6}{50^{10}}(49+10)$
$=\frac{59(49^9)}{(50^{10})}$

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Question 892 Marks

Let A and B be two independent events such that $P(A) = p_1$ and $P(B) = p_2$. Describe in words the events whose probabilities are:
$p_1p_2$.

Answer

As, $p_1p_2 = P(A) \times P(B)$
And, A and B are independent events.
i.e., $\text{P(A)}\times\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, $\text{P}(\text{A}\cap\text{B})=\text{P}_1\text{P}_2$
Hence, $p_1p_2$ = P(A and B occur).

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Question 902 Marks

Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that,

The youngest is a girl.
At least one is a girl?

Answer

Let a and g rapresent the boyy and the girl child respectively. if a family has two children, the samplw space will be
S = {(b, b), (b, g), (g, b), (g, g)}
Let a be the event that both children are girls.
$\therefore\ \text{A}=\big\{(\text{g},\text{g})\big\}$

Let B be the event that the youngest child is a girl.

$\therefore\text{B}=\big[(\text{b},\text{g}),(\text{g},\text{g})\big]$

$\Rightarrow\ \text{A}\cap\text{B}=\big\{(\text{g},\text{g})\big\}$

$\therefore\text{P(B)}=\frac{2}{4}=\frac{1}{2}$

$\text{P}(\text{A}\cap\text{B})=\frac{1}{4}$

The conditional probability that both are girls, given that the youngest child is a girl, is given by p (A | B).

$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$

Therefore, the required probability is $\frac{1}{2}$.

Let C be the event that at least one child is a girl.

$\therefore\text{C}=\big\{(\text{b},\text{g}),(\text{g},\text{b}),(\text{g},\text{g})\big\}$

$\Rightarrow\text{A}\cap\text{C}=\big\{\text{g},\text{g}\big\}$

$\Rightarrow\text{P(C)}=\frac{3}{4}$

$\text{P}(\text{A}\cap\text{C})=\frac{1}{4}$

The conditional probability that both are girls, given that at least one child a girl, is given by P(A|C).

$\therefore \text{P}(\text{A}|\text{C})=\frac{\text{P}(\text{A}\cap\text{C})}{\text{P}(\text{C})}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

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Question 912 Marks

The probability distribution of random variable X is given below:

$\text{X}$	$0$	$1$	$2$	$3$
$\text{P}(\text{X})$	$\text{k}$	$\frac{\text{k}}{2}$	$\frac{\text{k}}{4}$	$\frac{\text{k}}{8}$

Determine $\text{P}(\text{X}\leq2)$ and $\text{P}(\text{X}>2)$

Answer

$\text{P}(\text{X}\leq2)$
$=\text{P}(0)+\text{P}(1)+\text{P}(2)$
$=\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}$
$=\frac{8}{15}+\frac{8}{30}+\frac{8}{60}$
$=\frac{14}{15}$
$\text{P}(\text{X}>2)=\text{P}(3)=\frac{\text{k}}{8}=\frac{1}{15}$

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Question 922 Marks

The probability distribution of random variable X is given below:

$\text{X}$	$0$	$1$	$2$	$3$
$\text{P}(\text{X})$	$\text{k}$	$\frac{\text{k}}{2}$	$\frac{\text{k}}{4}$	$\frac{\text{k}}{8}$

Determine the value of k.

Answer

We know that,
$\text{P}(0)+\text{P}(1)+\text{P}(2)+\text{P}(3)=1$
$\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}+\frac{\text{k}}{8}=1$
$15\text{k}=8$
$\text{k}=\frac{8}{15}$

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Question 932 Marks

X is taking up subjects - Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets,
Grade A in no subject.

Answer

Given,
Probability of getting A grade in mathematics (m) = 0.2
⇒ P(m) = 0.2
Probability of getting A grade in physics (p) = 0.3
⇒ P(p) = 0.3
Probability of getting A grade in chemistry (P) = 0.5
⇒ P(C) = 0.5
P(Getting A in no subject)
$=\text{P}(\overline{\text{m}}\cap\overline{\text{p}}\cap\overline{\text{c}}\big)$
$=\text{P}(\overline{\text{m}})+\text{P}(\overline{\text{p}})+\text{P}(\overline{\text{c}}\big)$
$=\big(1-\text{P(m)}\big)\big(1-\text{P(p)}\big)\big(1-\text{P(c)}\big)$
$=(1-0.2)(1-0.3)(1-0.5)$
$=(0.8)(0.7)(0.5)$
$=0.28$
Required probability = 0.28

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Question 942 Marks

Fatima and John appear in an interview for two vacancies for the same post. The probability of Fatima's selection is $\frac{1}{7}$ and that of John's selection is $\frac{1}{5}$. What is the probability that,
None of them will be selected?

Answer

Given,
Probability of Fatima's (F) selection $=\frac{1}{7}$
$\text{P(F)}=\frac{1}{7}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{6}{7}$
Probability of John's (J) selection $=\frac{1}{5}$
$\text{P(F)}=\frac{1}{5}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{4}{5}$
P(None of them selected)
$=\text{P}(\overline{\text{F}}\cap\overline{\text{J}})$
$=\text{P}(\overline{\text{F}})+\text{P}(\overline{\text{J}})$
$=\frac{6}{7}\times\frac{4}{5}$
$=\frac{24}{35}$
Required probabilty $=\frac{24}{35}$

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Question 952 Marks

The probability distribution function oif a random variable X is given by

$X_i$	0	1	2
$P_i$	$3c^3$	$4c - 10c^2$	$5c - 1$

Where c > 0
Find: $\text{P}(1<\text{X}\leq2)$

Answer

$\text{P}(1<\text{X}\leq2)$
$=\text{P}(\text{X}=2)$
$=5\text{c}-1$
$=\frac{5}{3}-1$
$=\frac{5-3}{3}$
$=\frac{2}{3}$

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Question 962 Marks

One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
E: ‘the card drawn is black’
F: ‘the card drawn is a king’

Answer

In a deck of 52 cards, 26 cards are black and 4 cards are kings.
$\therefore$ P(E) = P(the card drawn is black) $=\frac{26}{52}=\frac{1}{2}$
$\therefore$ P(F) = P(the card drawn is a king) $=\frac{4}{52}=\frac{1}{13}$
In the pack of 52 cards, 2 cards are black as well as kings.
$\therefore$ P (EF) = P(the card drawn is a black king) $=\frac{2}{52}=\frac{1}{26}$
$\text{P}(\text{E})\times\text{P}(\text{F})=\frac{1}{2}\cdot\frac{1}{13}=\frac{1}{26}=\text{P}(\text{EF})$
Therefore, the given events E and F are independent.

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Question 972 Marks

Answer

Given,
Probability of getting A grade in mathematics (m) = 0.2
⇒ P(m) = 0.2
Probability of getting A grade in physics (p) = 0.3
⇒ P(p) = 0.3
Probability of getting A grade in chemistry (P) = 0.5
⇒ P(C) = 0.5
P(Getting A grade in all subjects)
$=\text{P}(\text{m}\cap\text{P}\cap\text{C})$
= P(m) + P(p) + P(c)
= 0.2 × 0.3 × 0.5
= 0.3
Required probability = 0.03

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Question 982 Marks

A husband and wife appear in an interview for two vacancies for the same post. The probability of husband's selection is $\frac{1}{7}$ and that of wife's selection is $\frac{1}{5}$. What is the probability that,
Both of them will be selected?

Answer

Given,
Probability of Husband's (H) selection $=\frac{1}{7}$
$\text{P(H)}=\frac{1}{7}$
Probability of wife's (W) selection $=\frac{1}{5}$
$\text{P(W)}=\frac{1}{5}$
P(Both of them will be selelcted)
$=(\text{H}\cap\text{W})$
$=\text{P(H)}\text{ P(W)}$
$=\frac{1}{7}\times\frac{1}{5}$
$=\frac{1}{35}$
Required probability $=\frac{1}{35}$

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Question 992 Marks

The mean of a binomial distribution is $10$ and its standard deviation is $2$; write the value of q.

Answer

Mean of the binomial distribution, i.e. $\text{np}=10$
$Variance = (Standard\ deviation)^2$, i.e. $\text{npq}=4$
$\therefore\text{q}=\frac{\text{Variance}}{\text{Mean}}=0.4$

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Question 1002 Marks

If A and B are independent events, then write expression for P(exactly one of A, B occurs).

Answer

As, A and B are independent events.
So, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}\ .....\text{(i)}$
P(exactly one of A, B occurs) = P(only A) + P(only B)
$=\big[\text{P(A)}-\text{P}(\text{A}\cap\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$=\big[\text{P(A)}-\text{P}(\text{A})\times\text{P}(\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A})\times\text{P}(\text{B})\big]$ [Using (i)]
$=\text{P(A)}\times\big[1-\text{P(B)}\big]+\text{P(B)}\times\big[1-\text{P(A)}\big]$
$=\text{P(A)}\times\text{P}(\overline{\text{B}})+\text{P(B)}\times\text{P}(\overline{\text{A}})$

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Maths 2 Marks Questions