3 Marks

Question 13 Marks

Find the mean and standard deviation of the following probability distributions:

$\text{x}_\text{i}$	$0$	$1$	$2$	$3$	$4$	$5$
$\text{p}_\text{i}$	$\frac{1}{6}$	$\frac{5}{18}$	$\frac{2}{9}$	$\frac{1}{6}$	$\frac{1}{9}$	$\frac{1}{18}$

Answer

$\text{x}_\text{i}$	$\text{p}_\text{i}$	$\text{p}_\text{i}\text{x}_\text{i}$	$\text{p}_\text{i}\text{x}_\text{i}^2$
$0$	$\frac{1}{6}$	$0$	$0$
$1$	$\frac{5}{18}$	$\frac{5}{18}$	$\frac{5}{18}$
$2$	$\frac{2}{9}$	$\frac{4}{9}$	$\frac{8}{9}$
$3$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{2}$
$4$	$\frac{1}{9}$	$\frac{4}{9}$	$\frac{16}{9}$
$5$	$\frac{1}{18}$	$\frac{5}{18}$	$\frac{25}{18}$
		$\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{35}{18}$	$\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{35}{6}$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{35}{18}$

Variance $\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2=\frac{35}{6}-\Big(\frac{35}{18}\Big)^2=\frac{665}{324}$

Standard Deviation $=\sqrt{\text{Variation}}=\frac{\sqrt{665}}{18}$

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Question 23 Marks

Find the mean and standard deviation of the following probability distributions:

x_i	-2	-1	0	1	2
p_i	0.1	0.2	0.4	0.2	0.1

Answer

x_i	p_i	x_ip_i	x_i²p_i
-2	0.1	-0.2	0.4
-1	0.2	-0.2	0.2
0	0.4	0	0
1	0.2	0.2	0.2
2	0.1	0.2	0.4
		$\sum\text{xp}=0$	$\sum\text{x}^2\text{p=1.2}$

Mean $=\sum\text{xp}$

Mean $=0$

Standard Deviation $=\sqrt{\sum\text{x}^2\text{p}-(\text{mean})^2}$

$=\sqrt{(1.2)^2-(0)^2}$

$=1.095$

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Question 33 Marks

Find the mean and standard deviation of the following probability distributions:

$\text{x}_\text{i}$	$-5$	$-4$	$1$	$2$
$\text{p}_\text{i}$	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{2}$	$\frac{1}{8}$

Answer

$\text{x}_\text{i}$	$\text{p}_\text{i}$	$\text{p}_\text{i}\text{x}_\text{i}$	$\text{p}_\text{i}\text{x}_\text{i}^2$
$-5$	$\frac{1}{4}$	$-\frac{5}{4}$	$\frac{25}{4}$
$-4$	$\frac{1}{8}$	$-\frac{4}{8}$	$\frac{16}{8}$
$1$	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
$2$	$\frac{1}{8}$	$\frac{2}{8}$	$\frac{4}{8}$
		$\sum\text{p}_\text{i}\text{x}_\text{i}=-1$	$\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{74}{8}$

$\sum\text{p}_\text{i}\text{x}_\text{i}=-1$

Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$

$=\frac{74}{8}-(-1)^2$

$=9.25-1$

$=8.25$

Step Deviation $=\sqrt{\text{Variance}}$

$=\sqrt{8.25}$

$=2.872$

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Question 43 Marks

Find the mean and standard deviation of the following probability distributions:

x_i	-3	-1	0	1	3
p_i	0.05	0.45	0.20	0.25	0.05

Answer

x_i	p_i	x_ip_i	x_i²p_i
-3	0.05	-0.15	0.45
-1	0.45	-0.45	0.45
0	0.20	0	0
1	0.25	0.25	0.25
3	0.05	0.15	0.45
		$\sum\text{xp}=-0.2$	$\sum\text{x}^2\text{p=1.6}$

Mean $=\sum\text{xp}$

Mean $=0.2$

Standard Deviation $=\sqrt{\sum\text{x}^2\text{p}-(\text{mean})^2}$

$=\sqrt{(1.6)-(-0.2)^2}$

$=\sqrt{1.6-0.04}$

$=\sqrt{1.56}$

Standard Deviation $=1.249$

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Question 53 Marks

Find the mean and standard deviation of the following probability distributions:

x_i	-1	0	1	2	3
p_i	0.3	0.1	0.1	0.3	0.2

Answer

x_i	p_i	p_ix_i	p_ix_i²
-1	0.3	-0.3	0.3
0	0.1	0	0
1	0.1	0.1	0.1
2	0.3	0.6	0.2
3	0.2	0.6	1.8
		$\sum\text{p}_\text{i}\text{x}_\text{i}=1$	$\sum\text{p}_\text{i}\text{x}_\text{i}^2=3.4$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=1$

Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$

$3.4-1$

$=2.4$

Step Deviation $=\sqrt{\text{Variance}}$

$=\sqrt{2.4}$

$=1.549$

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Question 63 Marks

Find the mean and standard deviation of the following probability distributions:

x_i	1	2	3	4
p_i	0.4	0.3	0.2	0.1

Answer

x_i	p_i	p_ix_i	p_ix_i²
1	0.3	0.4	0.4
2	0.1	0.6	1.2
3	0.1	0.6	1.8
4	0.3	0.4	1.6
		$\sum\text{p}_\text{i}\text{x}_\text{i}=2$	$\sum\text{p}_\text{i}\text{x}_\text{i}^2=5$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=2$

Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$

$=5-2^2$

$=1$

Step Deviation $=\sqrt{\text{Variance}}$

$=\sqrt{1}$

$=1$

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Question 73 Marks

Find the mean and standard deviation of the following probability distributions:

x_i	0	1	3	5
p_i	0.2	0.5	0.2	0.1

Answer

x_i	p_i	p_ix_i	p_ix_i²
0	0.2	0	0
1	0.5	0.5	0.5
3	0.2	0.6	1.8
5	0.1	0.5	2.5
		$\sum\text{p}_\text{i}\text{x}_\text{i}=1.6$	$\sum\text{p}_\text{i}\text{x}_\text{i}^2=4.8$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=1.6$

Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$

$=4.8-1.6^2$

$=2.24$

Step Deviation $=\sqrt{\text{Variance}}$

$=\sqrt{2.24}$

$=1.497$

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Question 83 Marks

Find the mean and standard deviation of the following probability distributions:

x_i	1	2	3	4
p_i	0.4	0.1	0.2	0.3

Answer

x_i	p_i	x_ip_i	x_i²p_i
1	0.4	0.4	0.4
3	0.1	0.3	0.9
4	0.2	0.8	3.2
5	0.3	1.5	7.5
		$\sum\text{xp}=3$	$\sum\text{x}^2\text{p}=12$

Mean $=\sum\text{xp}=3$

Standard deviation $=\sqrt{\sum\text{x}^2\text{p}-(\text{mean})^2}$

$=\sqrt{12-(3)^2}$

$=\sqrt3$

Standard deviation = 1.732

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Question 93 Marks

Find the mean and standard deviation of the following probability distributions:

x_i	2	3	4
p_i	0.2	0.5	0.3

Answer

x_i	p_i	p_ix_i	p_ix_i²
2	0.2	0.4	0.8
3	0.5	1.5	4.5
4	0.3	1.2	4.8
		$\sum\text{p}_\text{i}\text{x}_\text{i}=3.1$	$\sum\text{p}_\text{i}\text{x}_\text{i}^2=10.1$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=3.1$

Variation $=\sum\text{p}_\text{i}\text{x}_\text{i}^2=\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2=10.1-(3.1)^2=0.49$

Standard deviation $=\sqrt{\text{Variance}}=0.7$

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Question 103 Marks

Five defective bolts are acciedently mixed with twenty good ones. If four bolts are drawn at random from this lot, find the probability distribution of the number of defective bolts.

Answer

Here, 5 defective and 20 non-defective bolts. Let X denote the number of defective bolts drawn out of four bolts drawn. So, X can have values 1, 2, 3, 4.
P(X = 0)
$=\frac{\text{}^{20}\text{C}_4}{\text{}^{25}\text{C}_4}$
$=\frac{4845}{12650}$
$=\frac{969}{2530}$
P(X = 1)
$=\frac{\text{}^{5}\text{C}_1\times\text{}^{20}\text{C}_3}{\text{}^{25}\text{C}_4}$
$=\frac{5700}{12650}$
$=\frac{114}{253}$
P(X = 2)
$=\frac{\text{}^{5}\text{C}_2\times\text{}^{20}\text{C}_2}{\text{}^{25}\text{C}_4}$
$=\frac{1900}{12650}$
$=\frac{38}{253}$
P(X = 1)
$=\frac{\text{}^{5}\text{C}_3\times\text{}^{20}\text{C}_1}{\text{}^{25}\text{C}_4}$
$=\frac{200}{12650}$
$=\frac{4}{253}$
P(X = 4)
$=\frac{\text{}^{5}\text{C}_4}{\text{}^{25}\text{C}_4}$
$=\frac{5}{12650}$
$=\frac{1}{2530}$

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Question 113 Marks

If the probability distribution of a random variable of X is given by

X = x_i:	1	2	3	4
P(X = x_i):	2k	4k	3k	k

Write tyhe value of k.

Answer

Here,

X = x_i:	1	2	3	4
P(X = x_i):	2k	4k	3k	k

Since, $\sum\text{P}(\text{X})=1$

$\Rightarrow\text{P}(1)+\text{P}(2)+\text{P}(3)+\text{P}(4)=1$

$\Rightarrow2\text{k}+4\text{k}+3\text{k}+\text{k}=1$

$\Rightarrow10\text{k}=1$

$\Rightarrow\text{k}=\frac{1}{10}$

$\Rightarrow\text{k}=0.1$

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Question 123 Marks

Two cards are drawn successively without replacement from well shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Answer

Two cards are drawn successively without replacement from a pack of 52 cards. Let X denote the number of aces drawn from pack out of 2 cards. Then, X can take the values 0, 1 and 2.
Now,
P(X = 0)
$=\frac{48}{52}\times\frac{47}{51}$
$=\frac{2256}{2652}$
$=\frac{188}{221}$
P(X = 1)
$=\frac{4}{52}\times\frac{48}{51}+\frac{48}{52}\times\frac{4}{51}$
$=\frac{384}{2652}$
$=\frac{32}{221}$
P(X = 2)
$=\frac{4}{52}\times\frac{3}{51}$
$=\frac{12}{2652}$
$=\frac{1}{221}$
So, required probability distribution is

$\text{X}:$	$0$	$1$	$2$
$\text{P}(\text{X}):$	$\frac{188}{221}$	$\frac{32}{221}$	$\frac{1}{221}$

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Question 133 Marks

Two cards are drawn from a well shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Answer

Let X denote number of aces in a sample of 2 cards drawn.
There are four aces in a pack of 52 cards.
So, X can have values 0, 1, 2
Now,
P(X = 0)
$=\frac{\text{ }^{48}\text{C}_2}{\text{}^{52}\text{C}_2}$
$=\frac{2256}{2652}$
$=\frac{188}{221}$
P(X = 1)
$=\frac{\text{}^4\text{C}_1\times\text{}^{48}\text{C}_1}{\text{}^{52}\text{C}_2}$
$=\frac{192}{1326}$
$=\frac{32}{221}$
P(X = 2)
$=\frac{\text{ }^{4}\text{C}_2}{\text{}^{52}\text{C}_2}$
$=\frac{6}{1326}$
$=\frac{1 }{221}$
So,

$\text{X}:$	$0$	$1$	$2$
$\text{P}(\text{X}):$	$\frac{188}{221}$	$\frac{32}{221}$	$\frac{1}{221}$

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Question 143 Marks

A bag contains 4 red and 6 black balls. Three balls are drawn at random. Find the probability distribution of the number of red balls.

Answer

A bag contains 4 red and 6 black balls. Three balls are drawn.
Let X denote number of red balls out of three drawn.
Then X = 0, 1, 2, 3.
So,
P(no red balls) = P(X = 0)
$=\frac{\text{}^6\text{C}_3}{\text{}^{10}\text{C}_3}$
$=\frac{20}{120}$
$=\frac{1}{6}$
P(one red balls) = P(X = 1)
$=\frac{\text{}^4\text{C}_1\times\text{}^6\text{C}_2}{\text{}^{10}\text{C}_3}$
$=\frac{60}{120}$
$=\frac{1}{2}$
P(two red balls) = P(X = 2)
$=\frac{\text{}^4\text{C}_2\times\text{}^6\text{C}_1}{\text{}^{10}\text{C}_3}$
$=\frac{36}{120}$
$=\frac{3}{10}$
Pall three red) = P(X = 3)
$=\frac{\text{}^4\text{C}_3}{\text{}^{10}\text{C}_3}$
$=\frac{4}{120}$
$=\frac{1}{30}$
The required probability distribution is

$\text{X}:$	$0$	$1$	$2$	$3$
$\text{P}(\text{X}):$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

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Question 153 Marks

Find the probability distribution of white balls drawn in a random draw of 3 balls without replacement, from a bag containing 4 white and 6 red balls.

Answer

Given bag have 4 white balls and 6 red balls. Let X denote the number of white balls out of 3 balls drawn without replacement.
So, X = 0, 1, 2, 3.
P(No white balls)
$=\frac{\text{}^6\text{C}_3}{\text{}^{10}\text{C}_3}$
$=\frac{6\times5\times4}{3\times2}\times\frac{3\times2}{10\times9\times8}$
$=\frac{5}{30}$
P(One white balls)
$=\frac{\text{}^4\text{C}_1\times\text{}^6\text{C}_2}{\text{}^{10}\text{C}_3}$
$=\frac{4\times6\times5}{2}\times\frac{3\times2}{10\times9\times8}$
$=\frac{15}{30}$
P(Two white balls)
$=\frac{\text{}^4\text{C}_2\times\text{}^6\text{C}_1}{\text{}^{10}\text{C}_3}$
$=\frac{4\times3}{2}\times\frac{6\times3\times2}{10\times9\times8}$
$=\frac{9}{30}$
P(Three white balls)
$=\frac{\text{}^4\text{C}_3}{\text{}^{10}\text{C}_3}$
$=\frac{4\times3\times2\times1}{10\times9\times8}$
$=\frac{1}{30}$
So,
Required probability distribution is

$\text{X}:$	$0$	$1$	$2$	$3$
$\text{P}(\text{X}):$	$\frac{5}{30}$	$\frac{15}{30}$	$\frac{9}{30}$	$\frac{1}{30}$

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Question 163 Marks

In roulette, Figure, the wheel has 13 numbers 0, 1, 2,...., 12 maked on equally spaced slots. A player sets Rs 10 on a given number. He recieves Rs 100 from the organiser of the game if the ball comes to rest in this slot; otherwise he gets nothing. If X denotes the players net gain/loss, Find E(X).

Answer

$\text{P}(\text{win})=\frac{1}{13}\Rightarrow\text{P}(\text{lose})=\frac{12}{13}$
He gains Rs 90 if he wins and loses Rs 10 if his number does not appear.
Let X denote tptal loss or gain, so,

$\text{X}:$	$90$	$-10$
$\text{P}(\text{X}):$	$\frac{1}{13}$	$\frac{12}{13}$
$\text{XP}:$	$\frac{90}{13}$	$\frac{-120}{13}$

$\text{E}(\text{X})=\sum\text{XP}$
$=\frac{90}{13}-\frac{120}{13}$
$\text{E}(\text{X})=-\frac{30}{13}$

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Question 173 Marks

Find the probability distribution of Y in two throws of two dice, where Y represents the number of times a total of 9 appears.

Answer

It is given that Y denotes the number of times a total of 9 appears on throwing the pair of dice.
When the dice is thrown 2 times, the possibility of getting a total of 9 is possible only for the given combination:
(3, 6) (4, 5) (5, 4) (6, 3)
So, the total number of outcomes is 36 and the total number of favourable outcomes is 4.
Probability of getting a total of $9=\frac{4}{36}=\frac{1}{9}$
Probability of not getting a total of $9=1-\frac{1}{9}=\frac{8}{9}$
If Y takes the values 0, 1 and 2. Then,
$\text{P}(\text{Y}=0)=\frac{8}{9}\times\frac{8}{9}=\frac{64}{81}$
$\text{P}(\text{Y}=1)=\frac{1}{9}\times\frac{8}{9}+\frac{8}{9}\times\frac{1}{9}=\frac{16}{81}$
$\text{P}(\text{Y}=2)=\frac{1}{9}\times\frac{1}{9}=\frac{1}{81}$
Thus, the probability distribution of X is given by

$\text{Y}$	$0$	$1$	$2$
$\text{P}(\text{Y})$	$\frac{64}{81}$	$\frac{16}{81}$	$\frac{1}{81}$

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Question 183 Marks

Let X represents the difference between the number of heads and the number of tails when a coin is tossed 6 times. What are possible values of X?

Answer

Given: X = Number of heads - Number of tails

Number of heads	Number of heads	Number of heads - Number of tails
0	6	-6
1	5	-4
2	4	-2
3	3	0
4	2	2
5	1	4
6	0	6

Therefore, the possible values of X are:
-6, -4, -2, 0, 2, 4, 6.

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Question 193 Marks

For what value of k the following distribution is a probability distributioin?

X = x_i	0	1	2	3
P(X = x_i)	2k⁴	3k² - 5k³	2k - 3k²	3k - 1

Answer

We know that the sum of the probabilities in a probability distribution is always 1.
Therefore,
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1
⇒ 2k⁴ + 3k² - 5k³ + 2k - 3k² + 3k - 1 = 1
⇒ 2k⁴ -5k³ + 5k = 2
⇒ 2k⁴ -5k³ + 5k - 2 = 0
⇒ (k - 1)(k - 2)(2k2 + k - 1) = 0
⇒ (k - 1)(k - 2)(2k - 1)(k + 1) = 0
$\Rightarrow\text{k}=-1,\frac{1}{2},1,2$
(Negleting -1, 1 and 2 as they give the value of probability negative or greater than 1)
$\therefore\ \text{k}=\frac{1}{2}.$

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Question 203 Marks

Four cards are drawn simultaneously from a well shuffled pack of 52 playing cards. Find the probability distribution of the number of aces.

Answer

Let X denote number of aces in a sample of 4 cards drawn from a well shuffled pack of 52 playing cards. Then, X can take values 0, 1, 2, 3 and 4.
Now,
$\text{P}(\text{X}=0)=\text{P}(\text{no ace})=\frac{\text{}^{48}\text{C}_4}{\text{}^{52}\text{C}_4}$
$\text{P}(\text{X}=1)=\text{P}(\text{1 ace})=\frac{\text{}^{4}\text{C}_1\times\text{}^{48}\text{C}_3}{\text{}^{52}\text{C}_4}$
$\text{P}(\text{X}=2)=\text{P}(\text{2 aces})=\frac{\text{}^{4}\text{C}_2\times\text{}^{48}\text{C}_2}{\text{}^{52}\text{C}_4}$
$\text{P}(\text{X}=3)=\text{P}(\text{3 aces})=\frac{\text{}^{4}\text{C}_3\times\text{}^{48}\text{C}_1}{\text{}^{52}\text{C}_4}$
$\text{P}(\text{X}=4)=\text{P}(\text{4 aces})=\frac{\text{}^{4}\text{C}_4}{\text{}^{52}\text{C}_4}$
Thus, the probability distribution of X is given by

$\text{X}$	$\text{P}(\text{X})$
$0$	$\frac{\text{}^{48}\text{C}_4}{\text{}^{52}\text{C}_4}$
$1$	$\frac{\text{}^{4}\text{C}_1\times\text{}^{48}\text{C}_3}{\text{}^{52}\text{C}_4}$
$2$	$\frac{\text{}^{4}\text{C}_2\times\text{}^{48}\text{C}_2}{\text{}^{52}\text{C}_4}$
$3$	$\frac{\text{}^{4}\text{C}_3\times\text{}^{48}\text{C}_1}{\text{}^{52}\text{C}_4}$
$4$	$\frac{\text{}^{4}\text{C}_4}{\text{}^{52}\text{C}_4}$

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Question 213 Marks

Find the probability distribution of the number of heads, when three coins are tossed.

Answer

Let X denote number of heads in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.
Now,
$\text{P}(\text{X=0})=\text{P}(\text{TTT})=\frac{1}{8},\text{P}(\text{X}=1)$ $=\text{P}(\text{HTT or TTH or THT})=\frac{3}{8}$
$\text{P}(\text{X=2})\text{P}(\text{HTH or THH or HHT})$ $\\=\frac{3}{8},\text{P}(\text{X}=3)=\text{P}(\text{HHH})=\frac{1}{8}$
Thus, the probability distribution of X is given by

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

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Question 223 Marks

Two cards are drawn successively with replacement from well shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Answer

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
P(X = 0)
= P(no ace)
$=\frac{48}{52}\times\frac{48}{52}$
$=\frac{12\times12}{13\times13}$
$=\frac{144}{169}$
P(X = 1)
= P(1 ace)
$=\frac{4}{52}\times\frac{48}{52}$
$=\frac{2\times12}{13\times13}$
$=\frac{24}{169}$
P(X = 2)
= P(2 aces)
$=\frac{4}{52}\times\frac{4}{52}$
$=\frac{1\times1}{13\times13}$
$=\frac{1}{169}$
Thus, the probability distribution of X is given by

$\text{X}$	$0$	$1$	$2$
$\text{P}(\text{X})$	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$

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Question 233 Marks

A random variable X has the following probability distribution:

Values of X:	0	1	2	3	4	5	6	7	8
P(X)	a	3a	5a	7a	9a	11a	13a	15a	17a

Determine:

$\text{P}(\text{X}<3),\text{P}(\text{X}\geq3),\text{P}(0<\text{X}<5).$

Answer

$\text{P}(\text{X}<3)=\text{P}(0)+\text{P}(1)+\text{P}(2)$
$=\text{a}+3\text{a}+5\text{a}$
$=9\text{a}$
$=9\Big(\frac{1}{81}\Big)$
$\therefore\ \text{P}(\text{X}<3)=\frac{1}{9}$
$\text{P}(\text{X}\geq3)=1-\text{P}(\text{X}<3)=1-\frac{1}{9}=\frac{8}{9}$
$\text{P}(0<\text{X}<5)=\text{P}(1)+\text{P}(2)+\text{P}(3)+\text{P}(4)$
$=3\text{a}+5\text{a}+7\text{a}+9\text{a} $
$=24\text{a}$
$=24\Big(\frac{1}{81}\Big)$
$\therefore\ \text{P}(0<\text{X}<5)=\frac{8}{27}$

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Question 243 Marks

From a lot of 10 bulbs, which includes 3 defectives, a sample of two bulbs is drawn at random. Find the probability distribution of the number of defective bulbs.

Answer

Let X denote the number of defective bulbs in a sample of 2 bulbs drawn from a lot of 10 bulbs containing 3 defectives and 7 non defectives. Then X can take values 0, 1, 2.
Now,
P(X = 0) = P(no defective bulb) $=\frac{\text{}^{7}\text{C}_2}{\text{}^{10}\text{C}_2}=\frac{7}{15}$
P(X = 1) = P(1 defective bulb) $=\frac{\text{}^{3}\text{C}_1\times\text{}^{7}\text{C}_1}{\text{}^{10}\text{C}_2}=\frac{7}{15}$
P(X = 2) = P(2 defective bulbs) $=\frac{\text{}^{3}\text{C}_2}{\text{}^{10}\text{C}_2}=\frac{1}{15}$
Thus, the probability distribution of X is given below,

$\text{X}$	$0$	$1$	$2$
$\text{P}(\text{X})$	$\frac{7}{15}$	$\frac{7}{15}$	$\frac{1}{15}$

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Question 253 Marks

A box contains 13 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one without replacement, from the box. Find the probability distribution of the number of defective bulbs.

Answer

Let X denote the number of defective bulbs in a sample of 3 bulbs drawn from a bag containing 13 bulbs. 5 bulbs in the bag turn out to be defective. Then, X can take the values 0, 1, 2 and 3.
Now,
P(X = 0) = P(no defective bulb) $=\frac{\text{}^{8}\text{C}_3}{\text{}^{13}\text{C}_3}=\frac{56}{286}=\frac{28}{143}$
P(X = 1) = P(1 defective bulb) $=\frac{\text{}^{5}\text{C}_1\times\text{}^{8}\text{C}_2}{\text{}^{13}\text{C}_3}=\frac{140}{286}=\frac{70}{143}$
P(X = 2) = P(2 defecive bulbs) $=\frac{\text{}^{5}\text{C}_1\times\text{}^{8}\text{C}_1}{\text{}^{13}\text{C}_3}=\frac{80}{286}=\frac{40}{143}$
P(X = 2) = P(2 defecive bulbs) $=\frac{\text{}^5\text{C}_3}{\text{}^{13}\text{C}_3}=\frac{10}{286}=\frac{5}{143}$
Thus, the probability distribution of X is given by

$\text{X}$	$0$	$1$	$2$	$3$
$\text{P}(\text{X})$	$\frac{28}{143}$	$\frac{70}{143}$	$\frac{40}{143}$	$\frac{5}{143}$

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Question 263 Marks

An urn contain 5 red and 2 black balls. Two balls rendomly selected. Let X represent the number of black ball. What are the possible values of X. Is X a random variable?

Answer

Urn has 5 red and 2 black balls. 2 balls are rendomly selected.

Here, X denote the numbers of black balls.

So, possible values of X = 0, 1, 2

$\text{P}(\text{x}=0)=\text{P}\big(\overline{\text{B}}_1\big)\times\text{P}\big(\overline{\text{B}}_2\big)$

$=\frac{5}{7}\times\frac{5}{7}$

$=\frac{25}{49}$

$\text{P}(\text{X}=1)=\text{P}(\text{B}_1)\text{P}\big(\overline{\text{B}}_2\big)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}(\text{B}_2)$

$=\frac{2}{7}\times\frac{5}{7}+\frac{5}{7}\times\frac{2}{7}$

$=\frac{20}{49}$

$\text{P}(\text{X}=2)=\text{P}(\text{B}_1)\text{P}(\text{B}_2)$

$=\frac{2}{7}\times\frac{2}{7}$

$=\frac{4}{49}$

Now,

$\text{P}(\text{X}=0)+\text{P}(\text{X}=1)+\text{P}(\text{X}=2)$

$=\frac{25}{49}+\frac{20}{49}+\frac{4}{49}$

$=\frac{49}{49}$

$=1$

So, $\sum\text{P}(\text{X})=1$

Therefore,

X is a random variable.

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Question 273 Marks

A fair die is tossed twice. If the number appearing on the top is less than 3, it is success. Find the probability distribution of number of successes.

Answer

Let X denote the event of getting number less than 3 (1 or 2) on throwing the die. Then, X can take the values 0, 1 and 2.
Now,
$\text{P}(\text{X}=0)=\frac{16}{36}=\frac{4}{9}$
$\text{P}(\text{X}=1)=\frac{16}{36}=\frac{4}{9}$
$\text{P}(\text{X}=2)=\frac{4}{36}=\frac{1}{9}$
Thus, the probability distribution of X is given by

$\text{X}$	$0$	$1$	$2$
$\text{P}(\text{X})$	$\frac{4}{9}$	$\frac{4}{9}$	$\frac{1}{9}$

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Question 283 Marks

Let X be a random variable which assumes values x₁, x₂, x₃, x₄ such that 2P(X = x₁) = 3P(X = x₂) = P(X = x₃) = 5P(X = x₄). Find the probability distribution of X.

Answer

Here,

2P(x₁) = 3P(X₂) = P(x₃) = 5P(x₄)

Let P(x₃) = a

$2\text{P}(\text{x}_1)=\text{P}(\text{x}_3)\Rightarrow\text{P}(\text{x}_1)=\frac{\text{a}}{2}$

$3\text{P}(\text{x}_2)=\text{P}(\text{x}_3)\Rightarrow\text{P}(\text{x}_2)=\frac{\text{a}}{3}$

$5\text{P}(\text{x}_4)=\text{P}(\text{x}_3)\Rightarrow\text{P}(\text{x}_4)=\frac{\text{a}}{5}$

Since, $\text{P}(\text{x}_1)+\text{P}(\text{x}_2)+\text{P}(\text{x}_3)+\text{P}(\text{x}_4)=1$

$\Rightarrow\frac{\text{a}}{2}+\frac{\text{a}}{3}+\frac{\text{a}}{1}+\frac{\text{a}}{5}=1$

$\Rightarrow\frac{15\text{a}+10\text{a}+30\text{a}+6\text{a}}{30}=1$

$\Rightarrow61\text{a}=30$

$\Rightarrow\text{a}=\frac{30}{61}$

So,

$\text{X}:$	$\text{x}_1$	$\text{x}_2$	$\text{x}_3$	$\text{x}_4$
$\text{P}(\text{X}):$	$\frac{15}{61}$	$\frac{10}{61}$	$\frac{30}{61}$	$\frac{6}{61}$

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Question 293 Marks

A fair die is tossed. Let X denote twise the number appearing. Find the probability distribution, mean and variance of X.

Answer

$\text{x}_\text{i}$	$\text{p}_\text{i}$	$\text{p}_\text{i}\text{x}_\text{i}$	$\text{p}_\text{i}\text{x}_\text{i}^2$
$2$	$\frac{1}{6}$	$\frac{2}{6}$	$\frac{4}{6}$
$4$	$\frac{1}{6}$	$\frac{4}{6}$	$\frac{16}{6}$
$6$	$\frac{1}{6}$	$\frac{6}{6}$	$\frac{36}{6}$
$8$	$\frac{1}{6}$	$\frac{8}{6}$	$\frac{64}{6}$
$10$	$\frac{1}{6}$	$\frac{10}{6}$	$\frac{100}{6}$
$12$	$\frac{1}{6}$	$\frac{12}{6}$	$\frac{144}{6}$
		$\sum\text{x}_\text{i}\text{p}_\text{i}=7$	$\sum\text{x}_\text{i}\text{p}_\text{i}^2=\frac{364}{6}$

Mean $=\sum\text{x}_\text{i}\text{p}_\text{i}=7$

Variance $=\sum\text{x}_\text{i}\text{p}_\text{i}^2-(\text{Mean})^2$

$=60.7-49$

$=11.7$

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