ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.7

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.7

Question 1.
A boy throws a coin. He is to receive two rupees for getting a tail and ten rupees for getting a head. Find his expectation (mean receipt).
Answer:
In a single throw of coin, S = {H, T}
Let X (in Rs.) denotes the amount that the man gets
X = 2; P(X = 2) = P (T) = \(\frac{1}{2}\)
X = 10; P(H) = \(\frac{1}{2}\) =P(X = 10)
Mean = E (X) = Σp_ix_i = 2 × \(\frac{1}{2}\) + 10 × \(\frac{1}{2}\) = Rs. 6

Question 2.
A game at a school fair costs Rs. 10, with a prize of Rs. 50. If the probability of winning is \(\frac{1}{10}\) is it a fair game?
Answer:
Since the probability of winning the game = \(\frac{1}{10}\) = 0.1 < 0-5 the given game is not fair.

Question 3.
In a certain village, families are strictly limited to two children. The probability distribution of the number of children of any given individual is as follows :

Find the mean number of children.
Answer:
Given

Mean number of children = µ = E (X)
= \(\sum_{i=0}^2 x_i p\left(x_i\right)\)
= 0 × \(\frac{1}{10}\) + 1 × \(\frac{1}{2}\) + 2 × \(\frac{2}{5}\)
= \(\frac{1}{2}+\frac{4}{5}=\frac{5+8}{10}=\frac{13}{10}\)
= 1.3

Question 4.
Find the mean and the variance for the following probability distribution :

Answer:
Mean = µ = Σp_ix_i
= 0 × 0 1 + 1 × 0.25 + 2 × 0.3 + 3 × 0.2 + 4 × 0.15
= 0.25 + 0.6 + 0.6 + 0.6
= 2.05
Variance = Σp_ix_i – µ²
= (0.2 × 0.1 + 12 × .25 + 22 × 0.3 + 32 × 0.2 + 42 × 0.15) – (2.05)²
= (0 + 0.25 + 1.2 + 1.8 + 2.4) – (2.05)²
= 5.65 – 4.2025
= 1.4475

Question 5.
Find the mean µ and variance σ² for the following probability distribution : .

Answer:
The table of values is given as under:

Mean = µ = Σp_ix_i = \(\frac{6}{5}\) = 1.2
and variance = Σp_ix_i² – µ²
= 2 – \(\left(\frac{6}{5}\right)^2\)
= 2 – 1.44
= 0.56

Question 6.
A random variable X has the following probability distribution :

(i) Find the value of k.
(ii) Calculate the mean and the variance of the distribution.
Answer:
Σp_i = 1
⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1
⇒ 0.6 + 4k = 1
⇒ 4k = 1 – 0.6 = 0.4
⇒ k = \(\frac{0 \cdot 4}{4}\) = 0.1
Mean = µ = Σx_ip_i = (- 2) × 0.1 – 1 × k + 0 × 0.2 +1 × 2k + 2 × 0.3 + 3 × k
= – 0.2 – k + 0 + 2k + 0.6 + 3k
= 0.4 + 4k
= 0.4 + 4 × 0.1 = 0.8
Σx_ip_i = 4 × 0.1 + 1 × k+ 0 × 0-2 + 1 × 2k + 4 × 0-3 + 9 × k
= 0.4 + k + 0 + 2k + 1.2 + 9k
= 1.6 + 12k
= 1.6 + 12 × 0.1
= 1.6 + 1.2
= 2.8
Variance = σ² = Σx_ip_i – µ²
= 2.8 – (0.8)²
= 2.8 – 0.64
= 2.16

Question 7.
The probability distribution of a random variable X is given as below :

(i) Find the value of k.
(ii) Determine the mean of the distribution. [NCERT Exemplar]
Answer:
(i) Since the sum of all probabilities of probability distribution be equal to 1.
P (X = 0.5) + P (X = 1) + P (X = 1.5) + P (X = 2) = 1
⇒ k + k² + 2k² + k = 1
⇒ 3 k² + 2k – 1 = 0
⇒ (k + 1)(3k – 1) = 0
⇒ k = – 1,1/3
When k = – 1, P (X = 0-5) = – 1, which is not possible
since 0 < P (X) < 1
Thus k = \(\frac{1}{3}\)

(ii) Mean = Σp_ix_i
= [0.5 × k + 1 × k² + 1.5 x 2k² + 2 × k]
= 4k² + 2.5k
= \(\frac{4}{9}+\frac{2.5}{3}=\frac{4+7 \cdot 5}{9}=\frac{11.5}{9}\)
Thus,Mean = \(\frac{115}{90}=\frac{23}{18}\)

Question 8.
A discrete random variable X has the following probabilåty dbtribntion:

(i) Find the value of C
(ii) Find the mean of the distribution.
(iii) If Σp_ix_i ² = 14 find the variance of the distribution.
Answer:
(i) Since ΣP(X) = 2
⇒ 4C² + 3C² + 2C² + C² + C = 1
⇒ 10 C² + 3C – 1 = 0
⇒ C = \(\frac{-3 \pm \sqrt{9+40}}{20}=\frac{-3 \pm 7}{20}=\frac{1}{5},-\frac{1}{2}\)
but C > 0 ∴ C = \(\frac{1}{5}\)

(ii) ΣX P(X) = Mean = µ
⇒ µ = 0 × 4C² + 1 × 3C² + 2 2C² + 3 × C² + 4 × C + 5 × 2C²
= 10C² + 4C + 10C
= 10 × \(\frac{1}{25}+\frac{14}{5}=\frac{80}{25}=\frac{16}{5}\)

(iii) σ² = Σp_ix_i² – (Σp_ix_i)²
= 14 – µ²
= 14 – \(\left(\frac{16}{5}\right)^2\)
= \(\frac{350-256}{25}=\frac{94}{25}\)

Question 8(Old).
A random variable X has the following probability distribution :

Answer:
Since X be random variable I P (X) = 1
⇒ C + 2C + 2C + 3C + C² + 2C² + 7C² + C = 1
⇒ 10C² + 9C – 1 =0
⇒ (C + 1) (10C – 1) = 0
⇒ C = – 1, \(\frac{1}{10}\)
When C = -1
Then P(0) = -1 which is not possible as 0 ≤ P(X) ≤ 1
Thus C = \(\frac{1}{10}\)

Also Mean = ΣX P(X) = 0 + 2C + 4C + 9C + 4C2 + 10C² + 42C² + 6C²
= 56C² + 21C
= \(\frac{56}{100}+\frac{21}{10}=\frac{56+210}{100}\)
= \(\frac{266}{100}\)
= 2.66
or Mean = \(\frac{133}{50}\)

Question 9.
The probability distribution of a random variable X is given as below :

where k is a constant. Calculate
(i) the value of k.
(ii) E (X)
(iii) Standard deviation of X. (NCERT Example)
Answer:
Given

Since X be the fandom variable Σp_i = 1
⇒ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + 0 = 1
⇒ 2k + 3k + 4k + 5k + 10A + 12k + 14k = 1
⇒ 50k = 1
⇒ k = \(\frac{1}{50}\)
(ii)

E(X) = µ = Σp_ix_i = 1 × 2k + 2 × 3k + 4 × 5k + 5 × 10k + 6 × 12k + 7 × 14k + 8 × 0 …………. + 12 × 0
= 2k + 6k+ 20k + 50k + 12k + 98 k
= 260k
= \(\frac{260}{50}\) = 5.2

(iii) Σx_i²p_i = 1² × 2k + 2² × 3k + 3² × 4k + 4² × 5k + 5² × 10k + 6² × 12k + 7² × 14k
= 2k + 12k + 36k + 80k + 250k + 432k + 686k
= 1498k = \(\frac{1498}{50}\) = 29.96

Variance = Σx_i²p_i – μ² = 29.96 – (5.2)²
= 29.96 – 27.07
= 2.92
S.D (X) = \(\sqrt{2.92}\) = 1.71

Question 10.
Find the mean and variance of number of heads in three tosses of a fair coin.
Answer:
Let X denotes the number of tails in three tosses of a coin.
X can take values 0, 1, 2, 3.
P(X = 0) = probability of getting no tail = P (HHH) = \(\frac{1}{8}\)
P(X = 1) = probability of getting one tail and 2 heads = P (THH, HTH, HHT) = \(\frac{3}{8}\)
P(X = 2) = probability of getting two tails and one head = P (TTH, THT, HTT) = \(\frac{3}{8}\)
P(X = 3) = prob. of getting 3 tails = P (TTT) = \(\frac{1}{8}\)
probability distribution of X is as follows :

∴ Mean = Σx_ip_i = \(\frac{12}{8}=\frac{3}{2}\) and Variance = Σx_ip_i² – (Mean)²
= 3 – \(\frac{9}{4}=\frac{3}{4}\)

Question 10(Old).
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 ^ and 20 years.’ One student is randomly chosen and his/her age recorded. What is the probability distribution of the random variable X ? Find the mean, variance and standard distribution of X. (NCERT)
Answer:
The probability distribution of X is given as under
P (X = 14) = \(\frac{2}{15}\) P(X = 15) = \(\frac{1}{15}\);
P(X = 16) = \(\frac{2}{15}\); P(X = 17) = \(\frac{3}{15}\)
P(X = 18) = \(\frac{1}{15}\); P(X = 19) = \(\frac{2}{15}\)
P (X = 20) = \(\frac{3}{20}\); P (X = 21) = \(\frac{1}{15}\)

Required probability distribution of X is given below:

∴ Mean = µ = ΣX P (X)
= 14 × \(\frac{2}{15}\) + 15 × \(\frac{1}{15}\) + 16 × \(\frac{2}{15}\)+ 17 × \(\frac{3}{15}\) + 18 × \(\frac{1}{15}\) + 19 × \(\frac{2}{15}\) + 20 × \(\frac{3}{15}\) + 21 × \(\frac{1}{15}\)
= \(\frac{1}{15}\) [28+ 15 + 32 + 51 + 18 + 38 + 60 + 21]
= \(\frac{263}{15}\)

variance = Σx_i²/p_i -µ²
= 142 × \(\frac{2}{15}\)+ 152 × \(\frac{1}{15}\)+ 162 × \(\frac{2}{15}\)+ 172 × \(\frac{3}{15}\) + 182 × \(\frac{1}{15}\) + 192 × \(\frac{2}{15}\) + 202 × \(\frac{3}{15}\) + 212 × \(\frac{1}{15}\) – \(\left(\frac{263}{15}\right)^2\)
= \(\frac{1}{225}\) [(392 + 225 + 512 + 867 + 324 + 722 + 1200 + 441) 15 – 69169]
= \(\frac{1076}{225}\)

Question 11.
Two cards are drawn simultaneously from a pack of 52 cards. Compute the mean and standard deviation of the number of kings. [CBSE 2008]
Answer:
Let X denotes the number of kings drawn in a random sample of two cards drawn from a deck of 52 cards. Then X can take values 0, 1, 2.
P(X = 0) = prob. of getting no king card = \(\frac{{ }^{48} C_2}{{ }^{52} C_2}=\frac{48 \times 47}{52 \times 51}=\frac{564}{663}\)
P (X = 1) = prob. of getting 1 king card and one other card = \(\frac{{ }^{48} C_2}{{ }^{52} C_2}=\frac{48 \times 47}{52 \times 51}=\frac{564}{663}\)
P (X = 2) = prob. of getting 2 king cards = \(\frac{{ }^4 C_2}{{ }^{52} C_2}=\frac{4 \times 3}{52 \times 51}=\frac{3}{663}\)

The Probability distribution of random variable X is given as under

∴ Mean = Σp_ix_i = \(\sqrt{\frac{400}{2873}}=\frac{20}{\sqrt{2873}}\)
Variance = Σp_ix_i² – (Mean)²

∴ Variance = \(\frac{108}{663}-\left(\frac{34}{221}\right)^2=\frac{400}{2873}\)
and S.D = σ = \(\sqrt{\frac{400}{2873}}=\frac{20}{\sqrt{2873}}\)

Question 12.
An urn contains 5 red and 2 black balls. Two balls are randomly drawn without replacement. Find the probability distribution of the black balls drawn.
Also, find the mean and variance of the black balls drawn.
Answer:
Now two balls are drawn without replacement is same is drawing two balls simultaneously.
Here total no. of balls = 5 + 2 = 7
Let X be the random variable denote the no. of black balls drawn in a draw of two balls.

Thus probability distribution of random variable X be given by :

∴ Mean = µ = Σp_ix_i
= 0 × \(\frac{10}{21}\) + 1 × \(\frac{10}{21}\) + 2 × \(\frac{1}{21}=\frac{12}{21}=\frac{4}{7}\)
and Variance = σ² = Σp_ix_i² – µ²
= 0² × \(\frac{10}{21}\) + 1² × \(\frac{10}{21}\) + 2² × \(\frac{1}{21}-\left(\frac{4}{7}\right)^2\)
= \(\frac{2}{3}-\frac{16}{49}=\frac{98-48}{147}=\frac{50}{147}\)

Question 13.
3 defective bulbs are mixed with 7 good ones. Find the probability distribution of the number of defective bulbs, if 3 bulbs are drawn at random. What is the average number of defective bulbs drawn ?
Answer:
Total no. of defective bulbs = 3 and no. of good bulbs = 7
∴ Total no. of bulbs = (3 + 7) = 10
Let X be the random variable denotes the no. of defective bulbs in a draw of 3 bulbs.
So X can takes values 0, 1, 2, 3.
P (X = 0) = P (drawing no defective bulb)
= P(drawing 3 good bulbs) = \(\frac{{ }^7 C_3}{{ }^{10} C_3}=\frac{7 \times 6 \times 5}{10 \times 9 \times 8}=\frac{7}{24}\)

P(X = 1) = P (drawing one defective and two
= \(\frac{{ }^3 C_1 \times{ }^7 C_2}{{ }^{10} C_3}=\frac{3 \times 7 \times 6 \times 6}{2 \times 10 \times 9 \times 8}=\frac{21}{40}\)

P (X = 2) = P (drawing two defective and one good bulb)
= \(\frac{{ }^3 C_2 \times{ }^7 C_1}{{ }^{10} C_3}=\frac{3 \times 7 \times 6}{10 \times 9 \times 8}=\frac{7}{40}\)

P (X = 3) = P (drawing three defective bulbs)
= \(\frac{{ }^3 C_3}{{ }^{10} C_3}=\frac{6}{10 \times 9 \times 8}=\frac{1}{120}\)

The probability distribution of X is given by

required average = µ = ΣX P (X) = 0 × \(\frac{7}{24}\)+ 1 × \(\frac{21}{40}\) + 2 × \(\frac{7}{40}\) + 3 × \(\frac{1}{120}\)
= 0 + \(\frac{21}{40}+\frac{14}{40}+\frac{1}{40}=\frac{36}{40}=\frac{9}{10}\)

Question 14.
2 bad eggs are accidentally mixed with 10 good ones. If three eggs are drawn at random, find the probability distribution of bad eggs drawn. Also find the mean and variance of the distribution.
Answer:
X denotes the number of bad eggs drawn in a random draw of 3 eggs with replacement from a lot containing two bad eggs and 10 good eggs.
So X can take values 0, 1, 2, 3.
P (X = 0) = prob. of getting no bad egg
= prob. of getting three good ones in three draws = \(\frac{{ }^{10} C_3}{{ }^{12} C_3}=\frac{6}{11}\)

P(X = 1) = prob. of getting one bad egg and two good ones in 3 draws = \(\frac{{ }^{10} C_3}{{ }^{12} C_3}=\frac{6}{11}\)

P(X = 2) = prob. of getting 2 bad eggs and one good on in three draws = \(\frac{{ }^{10} C_3}{{ }^{12} C_3}=\frac{6}{11}\)

The probability distribution of random variable X is given below :

Mean = 0 × \(\frac{6}{11}\) + 1 × \(\frac{9}{22}\) + 2 × \(\frac{1}{22}=\frac{11}{22}=\frac{1}{2}\)
Variance = 0² × \(\frac{6}{11}\) + 1² × \(\frac{9}{22}\) + 4 × \(\frac{1}{22}=\frac{11}{22}=\frac{1}{2}\)

Question 15.
A box contains 4 red and 5 black marbles. Find the probability distribution of the red marbles in a random draw of three marbles. Also find the mean and standard deviation of the distribution.
Answer:
Given, no. of red marbles in a box = 4
no. of back marbles in a box = 5
Total number of marbles in a box = 4 + 5 = 9 ,
Let X be the random variable denotes the no. of red marbles in a randoin draw of three marbles
∴ X takes values 0, 1, 2, 3.

The total no. of ways of choosing 3 marbles out of 9 be ⁹C₃.
P (X = 0) = Prob. (drawing no red bulbs) = P (drawing 3 black marbles)
= \(\frac{{ }^5 C_3}{{ }^9 C_3}=\frac{5 \times 4 \times 6}{2 \times 9 \times 8 \times 7}=\frac{5}{42}\)

P (X = 1) = \(\frac{{ }^5 C_3}{{ }^9 C_3}=\frac{5 \times 4 \times 6}{2 \times 9 \times 8 \times 7}=\frac{5}{42}\) [P (X = 1)= P (drawing 1 red and 2 black marbles]
= \(\frac{20}{42}\)

P (X = 2) = P (drawing 2 red and 1 black marble)
= \(\frac{{ }^4 \mathrm{C}_2 \times{ }^5 \mathrm{C}_1}{{ }^9 \mathrm{C}_3}=\frac{4 \times 3 \times 5 \times 6}{2 \times 9 \times 8 \times 7}=\frac{15}{42}\)

P (X = 3) = P (drawing 3 red marbles)
= \(\frac{{ }^4 C_3}{{ }^9 C_3}=\frac{4 \times 6}{9 \times 8 \times 7}=\frac{1}{21}\)

The probability distribution of X is given below :

Question 16.
From a lot of 6 items containing 2 defective items, a sample of 4 items is drawn at random (without replacement). If the random variable X denotes the number of defective items in the sample, find :
(i) the probability distribution of X.
(ii) the mean of the distribution.
(iii) the variance of the distribution.
Answer:
Given X be the random variable donates the number of defective items in the sample. Then X can take values 0, 1,2.
Total no. of drawing four items = ⁶C₄
given no. of defective items = 2
No. of good items = 6 – 2 = 4
P (X = 0) = prob. of getting no defective item = \(\frac{{ }^4 C_4}{{ }^6 C_4}=\frac{2}{30}=\frac{1}{15}\)

P(X = 1) = prob. of getting one defective item = \(\frac{{ }^2 C_1 \times{ }^4 C_3}{{ }^6 C_4}=\frac{2 \times 4 \times 2}{30}=\frac{8}{15}\)

P(X = 2) = prob. of getting two defective item = \(\frac{{ }^2 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2}{{ }^6 \mathrm{C}_4}=\frac{1 \times 4 \times 3 \times 2}{2 \times 6 \times 5}=\frac{6}{15}\)

Thus the probability distribution of X is given as under :

The table of values are given as under:

∴ Mean = Σp_ix_i = μ = \(\frac{20}{15}=\frac{4}{3}\)
Variable = σ² = Σp_ix_i² – μ²
= \(\frac{20}{15}=\frac{4}{3}\)
= \(\frac{96-80}{45}=\frac{16}{45}\)

Question 17.
Two positive numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and the variance of the distribution.
Answer:
The total number of ways of selecting two numbers (without replacement) out of first five positive integers = ⁵C₂ = \(\frac{5 \times 4}{2}\) = 10
So all the 10 outcomes are equally likely and Exhaustive and these outcomes are
1, 2 ; 1, 3 ; 1, 4 ; 1, 5 ; 2, 3 ; 2, 4 ; 2, 5 ; 3, 4 ; 3, 5 ; 4, 5.
Let X be the random variable denotes the larger of two numbers so X can take the values 2, 3, 4, 5
Since 1 is not larger than any of number from 1 to 5.

Larger of the number	No. of outcomes
2	1 {1, 2}
3	2 (1, 2), (2, 3)
4	3 (1, 4), (2, 4), (3, 4)
5	4 (1, 5), (2, 5), (3, 5), (4, 5)

Thus the probability distribution of X be :

Question 18.
Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the probability distribution of X. Also find the mean and the variance of the distribution.
Answer:
The number of ways of choosing two integers (without replacement) from given six positive integers = ⁶C₂ = \(\frac{6 \times 5}{2}\) = 15
Sample space for this experiment is given below
{2, 3 ; 2, 4 ; 2, 5 ; 2, 6 ; 2, 7 ; 3, 4 ; 3, 5 ; 3, 6 ; 3, 7 ; 4, 5 ; 4, 6 ; 4, 7 ; 5, 6 ; 5, 7 ; 6, 7}
all these outcomes are equally likely.
Since the random variable X denotes the larger of two numbers.
∴ X can take values, 3, 4, 5, 6 and 7.
Since 2 is not larger than any of given integers 2, 3, 4, 5, 6 and 7.

Larger of two numbers	No. of favourable outcomes
3	1
4	2
5	3
6	4
7	5

P(X = 3) = \(\frac{1}{15}\); P(X = 4) = \(\frac{2}{15}\) P(x = 5) = \(\frac{3}{15}\) P(X = 6) = \(\frac{4}{15}\); P(X = 7) = \(\frac{5}{15}\)

The required probability distribution of X is given below :

Question 19.
Three numbers are selected at random (without replacement) from the first six positive integers. Let X denote the largest of the three numbers obtained. Find the probability distribution of X. Also find the mean and the variance of the distribution.
Answer:
The total no. of ways of choosing three integers (without replacement) from given first six integers = ⁶C₃ = \(\frac{6 \times 5 \times 4}{6}\) = 20
Here S = {(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6)}
All these outcomes are equally likely. Since the random variable X denotes the larger of three numbers.
∴ X can takes values 3, 4, 5, 6.
Since 1, 2 are not larger than any of the given integers 3, 4, 5, 6.

Larger of three numbers	No. of favourable outcomes
3	1
4	3
5	6
6	10

∴ P(X = 3) = \(\frac{1}{20}\); P(X = 4)= \(\frac{3}{20}\); P(X = 5) = \(\frac{6}{20}\); P(X = 6)= \(\frac{10}{20}\)

The required probability distribution of X is given by

Question 20.
There are 4 cards numbered 1,3, 5 and 7, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and variance of X.
Answer:
The total number of ways of drawing two cards (without replacement) out of four given cards = ⁴C₂ = \(\frac{4 \times 3}{2}\) = 6
So all the 6 outcomes are equally likely and given as under : 1,3; 1,5; 1,7; 3, 5; 3, 7; 5, 7
Let X be the random variable denotes the sum of the numbers on two drawn cards.
∴ X can take values 4, 6, 8, 10, 12.

Sum of numbers X	No. of outcomes
4	1
6	1
8	2
10	1
12	1

∴ P (x = 4) = \(\frac{1}{6}\); P (x = 6) = \(\frac{1}{6}\); P (X = 8) = \(\frac{2}{6}\); P(X= 10) = \(\frac{1}{6}\)
and P (X = 12) = \(\frac{1}{6}\)
Thus probability distribution of X is given below :