Self Assessment and Revision on Theoritical Probability Distribution Class 12 OP Malhotra ISC Maths Solutions Ch-20. In this article you would learn full concept recap of Probability. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Self Assessment and Revision on Theoritical Probability Distribution Class 12 OP Malhotra  ISC Maths Solutions Ch-20

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-20	Theoritical Probability Distribution
Writer	OP Malhotra
Exe	Self Assessment and Revision

Self Assessment and Revision on Theoritical Probability Distribution

 OP Malhotra ISC Class 12 Maths Solutions

Ques-1. In a large collection of bulbs, 3 out of 5 are daffodils and the rest are tulips. If they are planted at random, calculate the probability that in a row of five plants:
(i) all are daffodils,
(ii) at least four are daffodils. 

Ans- Given: Probability of daffodils p = 3/5, tulips q = 2/5, number of plants n = 5

(i) Probability that all are daffodils:
P(X = 5) = ⁵C₅(3/5)⁵(2/5)⁰
= 1 × (243/3125) = 243/3125

(ii) Probability that at least 4 are daffodils:
P(X ≥ 4) = P(4) + P(5)
P(4) = ⁵C₄(3/5)⁴(2/5)
= 5 × (81/625) × (2/5) = 810/3125

P(5) = 243/3125

Total probability = (810 + 243)/3125 = 1053/3125

Ques-2. Mohan and Sohan play 12 games of chess. Mohan wins 6 games and Sohan wins 4 games. 2 games end in a draw. They agree to play 3 more games. Calculate the probability that out of these 3 games, 2 games end in a draw. (ISC 1993)

Ans- Probability of draw p = 2/12 = 1/6, probability of not draw q = 5/6, number of games n = 3

Required probability:
P(X = 2) = ³C₂(1/6)²(5/6)
= 3 × (1/36) × (5/6)
= 15/216 = 5/72

Ques-3. On an average, out of 12 games of chess played by A and B, A wins 6, B wins 4 and 2 games end in a tie. A and B play a tournament of 3 games. Calculate the probability that:
(i) at least 2 games end in a tie.
(ii) A and B win alternate games, no games end in a tie. 

Ans- Probabilities:
P(A wins) = 6/12 = 1/2
P(B wins) = 4/12 = 1/3
P(tie) = 2/12 = 1/6

(i) Probability that at least 2 games end in a tie:
P(X ≥ 2) = P(2) + P(3)

P(2) = ³C₂(1/6)²(5/6)
= 3 × (1/36) × (5/6) = 15/216

P(3) = (1/6)³ = 1/216

Total probability = (15 + 1)/216 = 16/216 = 2/27

(ii) Probability that A and B win alternate games and no game is a tie:
Possible sequences: A B A and B A B

P(ABA) = (1/2)(1/3)(1/2) = 1/12
P(BAB) = (1/3)(1/2)(1/3) = 1/18

Total probability = 1/12 + 1/18 = (3 + 2)/36 = 5/36

Ques-4. The probability density function y of a continuous variable x is given by y = x/k, 0 ≤ x ≤ 2, and y = 0 for all other values of x. Calculate the value of k and the probability that x < 1. 

Ans- Given: y = x/k for 0 ≤ x ≤ 2

Total probability = 1
∫₀² (x/k) dx = 1

(1/k) ∫₀² x dx = 1
(1/k)[x²/2]₀² = 1
(1/k)(4/2) = 1
2/k = 1 ⇒ k = 2

Now, P(x < 1):
P(x < 1) = ∫₀¹ (x/2) dx
= (1/2)[x²/2]₀¹
= (1/2)(1/2) = 1/4

Ques-5. A pair of dice is thrown 5 times. If getting a doublet is considered a success, find the probability of 2 successes. (ISC 1995)

Ans- Total outcomes when two dice are thrown = 36
Number of doublets = 6 ⇒ p = 6/36 = 1/6
q = 5/6, n = 5

P(X = 2) = ⁵C₂(1/6)²(5/6)³
= 10 × (1/36) × (125/216)
= 1250/7776
= 625/3888

Ques-6. Assume that on an average, 1 telephone out of 10 is busy. Six telephone numbers are randomly selected and called. Find the probability that four of them will be busy. (ISC 1997)

Ans- p = 1/10, q = 9/10, n = 6

P(X = 4) = ⁶C₄(1/10)⁴(9/10)²
= 15 × (1/10000) × (81/100)
= 1215/1000000
= 0.001215

Ques-7. Assume that half the population are consumers of chocolate, so that the chance of an individual being a consumer is 1/2. If 100 investigators interviewed 10 individuals each to see whether they are consumers, how many investigators would you expect to report that 3 individuals or less are consumers. 

Ans- p = 1/2, q = 1/2, n = 10

P(X ≤ 3) = [¹⁰C₀ + ¹⁰C₁ + ¹⁰C₂ + ¹⁰C₃] / 2¹⁰
= (1 + 10 + 45 + 120)/1024
= 176/1024

Expected number = 100 × (176/1024) ≈ 17

Ques-8. A and B play a game in which A’s chance of winning the game is 3/5. In a series of 6 games, find the probability that A will win at least 4 games. 

Ans- p = 3/5, q = 2/5, n = 6

P(X ≥ 4) = P(4) + P(5) + P(6)

P(4) = ⁶C₄(3/5)⁴(2/5)²
= 15 × (81/625) × (4/25)
= 4860/15625

P(5) = ⁶C₅(3/5)⁵(2/5)
= 6 × (243/3125) × (2/5)
= 2916/15625

P(6) = (3/5)^{6</sup = 729/15625}

Total = (4860 + 2916 + 729)/15625 = 8505/15625
= 1701/3125

Ques-9. A die is tossed three times. Getting a ‘3’ or a ‘5’ is considered a success. Find the probability of at least two successes.

Ans- Success outcomes = {3,5} ⇒ p = 2/6 = 1/3, q = 2/3, n = 3

P(X ≥ 2) = P(2) + P(3)

P(2) = ³C₂(1/3)²(2/3)
= 3 × (1/9) × (2/3) = 6/27

P(3) = (1/3)³ = 1/27

Total = 7/27

Ques-10. A coin is tossed 5 times. What is the probability of getting at least three heads?

Ans- n = 5, p = 1/2

P(X ≥ 3) = [⁵C₃ + ⁵C₄ + ⁵C₅] / 2⁵
= (10 + 5 + 1)/32
= 16/32 = 1/2

Ques-11. Four dice are thrown simultaneously. If the occurrence of an odd number in a single die is considered a success, find the probability of at most 2 successes.

Ans- Ek die par odd number (1,3,5) aane ki probability = 3/6 = 1/2 ⇒ p = 1/2
Even number ki probability q = 1/2, aur number of trials n = 4

“At most 2 successes” ka matlab hai X ≤ 2 (0,1,2 cases)

P(X ≤ 2) = P(0) + P(1) + P(2)

P(0) = ⁴C₀(1/2)⁰(1/2)⁴ = 1/16
P(1) = ⁴C₁(1/2)¹(1/2)³ = 4/16
P(2) = ⁴C₂(1/2)²(1/2)² = 6/16

Total probability = (1 + 4 + 6)/16 = 11/16

Ques-12. Assuming that on an average one telephone out of ten is busy, seven telephone numbers are randomly selected and called. Find the probability that three of them will be busy. (ISC 2007)

Ans- Ek telephone busy hone ki probability p = 1/10, free hone ki q = 9/10
Number of trials n = 7

Exactly 3 busy hone ki probability:

P(X = 3) = ⁷C₃(1/10)³(9/10)⁴

= 35 × (1/1000) × (6561/10000)
= 229635/10000000

Ques-13. In a binomial distribution, the sum of its mean and the variance is 1.8. Find the probability of two successes if the experiment is conducted 5 times. (ISC 2007)

Ans- Binomial distribution mein:
Mean = np
Variance = npq

Given: np + npq = 1.8
⇒ np(1 + q) = 1.8
⇒ np(2 − p) = 1.8 (kyunki q = 1 − p)

n = 5 given hai, toh try karte hain p = 0.2 (check):
np = 5 × 0.2 = 1
npq = 5 × 0.2 × 0.8 = 0.8
Sum = 1 + 0.8 = 1.8

Ab P(X = 2):
P(2) = ⁵C₂(0.2)²(0.8)³
= 10 × 0.04 × 0.512 = 0.2048

Ques-14. Eight coins are thrown simultaneously. (i) Show that the probability of getting at least 6 heads is 37/256. (ii) Find the probability of getting at least 3 heads. (ISC 2011, 2008)

Ans- Coin ke liye p = 1/2, n = 8

(i) At least 6 heads:
P(X ≥ 6) = P(6) + P(7) + P(8)

P(6) = ⁸C₆/256 = 28/256
P(7) = 8/256
P(8) = 1/256

Total = (28 + 8 + 1)/256 = 37/256

(ii) At least 3 heads:
P(X ≥ 3) = 1 − [P(0) + P(1) + P(2)]
= 1 − (1 + 8 + 28)/256
= 1 − 37/256 = 219/256

Ques-15. The mean and variance of a binomial distribution are 4 and 2 respectively. Find the probability of at least 6 successes. (ISC 2009)

Ans- Mean = np = 4
Variance = npq = 2

⇒ q = 2/4 = 1/2 ⇒ p = 1/2
⇒ n = 8

P(X ≥ 6) = P(6) + P(7) + P(8)

= [⁸C₆ + ⁸C₇ + ⁸C₈] / 256
= (28 + 8 + 1)/256 = 37/256

Ques-16. Five bad eggs are mixed with 10 good ones. If three eggs are drawn one by one with replacement, find the probability distribution of the number of good eggs drawn.

Ans- Good eggs ki probability p = 10/15 = 2/3
Bad eggs ki q = 1/3, n = 3

X = number of good eggs

P(0) = (1/3)³ = 1/27
P(1) = ³C₁(2/3)(1/3)² = 6/27
P(2) = ³C₂(2/3)²(1/3) = 12/27
P(3) = (2/3)³ = 8/27

X (No. of good eggs)	P(X)
0	1/27
1	6/27
2	12/27
3	8/27

Ques-17. 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. 

Ans- Total marbles = 9 (4 red + 5 black)
Number of draws n = 3 (without replacement)

Since drawing is without replacement, the distribution is hypergeometric.

P(X = x) = [⁴C_x × ⁵C_3−x] / ⁹C₃

Total ways = ⁹C₃ = 84

X (No. of red marbles)	P(X)
0	[4C0 × 5C3] / 84 = (1 × 10)/84 = 10/84 = 5/42
1	[4C1 × 5C2] / 84 = (4 × 10)/84 = 40/84 = 10/21
2	[4C2 × 5C1] / 84 = (6 × 5)/84 = 30/84 = 5/14
3	[4C3 × 5C0] / 84 = (4 × 1)/84 = 4/84 = 1/21

Total probability = 5/42 + 10/21 + 5/14 + 1/21 = 1

Mean (E(X)) = n × (number of red / total marbles)
= 3 × (4/9) = 4/3

Variance = n × (4/9) × (5/9) × [(9−3)/(9−1)]
= 3 × (4/9) × (5/9) × (6/8)
= 3 × 20/81 × 6/8
= 360/648 = 5/9

Standard Deviation = √(5/9) = √5 / 3

Ques-18. The probability that a bulb will fuse in 100 days is 0.05. Find probability that out of 5 bulbs: (i) none (ii) not more than one (iii) more than one (iv) at least one fuse.

Ans- p = 0.05, q = 0.95, n = 5

(i) None:
P(0) = (0.95)⁵

(ii) Not more than one:
P(0) + P(1)
= (0.95)⁵ + ⁵C₁(0.05)(0.95)⁴

(iii) More than one:
= 1 − [P(0) + P(1)]

(iv) At least one:
= 1 − (0.95)⁵

Ques-19. If the sum and product of the mean and variance of a binomial distribution are 1.8 and 0.8 respectively, find the distribution and probability of at least one success.

Ans- Let mean = μ and variance = σ²

Given:
μ + σ² = 1.8 …(1)
μ × σ² = 0.8 …(2)

From (1): σ² = 1.8 − μ

Substitute in (2):
μ(1.8 − μ) = 0.8

⇒ 1.8μ − μ² = 0.8
⇒ μ² − 1.8μ + 0.8 = 0
⇒ μ² − 1.8μ + 0.8 = 0

Solving quadratic:
μ = [1.8 ± √(1.8² − 4×0.8)] / 2
= [1.8 ± √(3.24 − 3.2)] / 2
= [1.8 ± √0.04] / 2
= [1.8 ± 0.2] / 2

So, μ = 1 or μ = 0.8

Now check valid case:
For binomial distribution:
Mean μ = np
Variance σ² = npq = np(1−p)

Case 1: μ = 1
Then σ² = 1.8 − 1 = 0.8

np = 1 …(i)
npq = 0.8 …(ii)

From (ii)/(i): q = 0.8 ⇒ p = 0.2

From (i): n × 0.2 = 1 ⇒ n = 5

Thus, distribution is:
n = 5, p = 0.2, q = 0.8

(Case μ = 0.8 gives non-integer n, so rejected)

Now probability of at least one success:

P(X ≥ 1) = 1 − P(0)
= 1 − (q)ⁿ
= 1 − (0.8)⁵
= 1 − 0.32768
= 0.67232

Ques-20. On dialling telephone numbers, assume 1 out of 5 is busy. Ten numbers are selected. Find probability that at least 3 are busy. (ISC 2014)

Ans- Probability that a number is busy = p = 1/5
Probability that a number is not busy = q = 4/5
Number of trials n = 10

Let X = number of busy telephone numbers.

We need P(X ≥ 3). It is easier to use the complement rule:

P(X ≥ 3) = 1 − [P(0) + P(1) + P(2)]

Using the binomial formula:
P(X = r) = ⁿC_r p^r q^n−r

P(0) = ¹⁰C₀(1/5)⁰(4/5)¹⁰ = (4/5)¹⁰

P(1) = ¹⁰C₁(1/5)(4/5)⁹ = 10(1/5)(4/5)⁹

P(2) = ¹⁰C₂(1/5)²(4/5)⁸ = 45(1/25)(4/5)⁸
Taking (4/5)⁸ common:

P(X ≥ 3) = 1 − (4/5)⁸ [ (4/5)² + 10(1/5)(4/5) + 45(1/25) ]
= 1 − (4/5)⁸ [ 16/25 + 40/25 + 45/25 ]
= 1 − (4/5)⁸ (101/25)

Ques-21. Five dice are thrown simultaneously. If occurrence of odd number is success, find probability of maximum successes. 

Ans- On a die, odd numbers are: {1, 3, 5} ⇒ total 3 outcomes
Total outcomes = 6

So, probability of success (odd number):
p = 3/6 = 1/2

Probability of failure (even number):
q = 1/2

Number of trials n = 5

“Maximum successes” means all 5 dice show odd numbers.
So, we need P(X = 5)

Using binomial formula:

P(X = r) = ⁿC_r p^r q^n−r
P(X = 5) = ⁵C₅ (1/2)⁵ (1/2)⁰
= 1 × (1/2)⁵
= 1/32

Ques-22. The difference between mean and variance of a binomial distribution is 1 and the difference of their squares is 11. Find the distribution. 

Ans- Let mean = μ and variance = σ²

For a binomial distribution:
μ = np and σ² = npq = np(1 − p)

Given:
μ − σ² = 1 …(1)
μ² − (σ²)² = 11 …(2)

From (2):
μ² − (σ²)² = (μ − σ²)(μ + σ²)

Using (1):
1 × (μ + σ²) = 11
⇒ μ + σ² = 11 …(3)

Now solve (1) and (3):
μ − σ² = 1
μ + σ² = 11

Adding:
2μ = 12 ⇒ μ = 6

Substitute in (1):
6 − σ² = 1 ⇒ σ² = 5

Now,
μ = np = 6 …(i)
σ² = npq = 5 …(ii)

Divide (ii) by (i):
q = 5/6 ⇒ p = 1 − 5/6 = 1/6

Substitute in (i):
n × (1/6) = 6 ⇒ n = 36

Thus, the binomial distribution is:
n = 36, p = 1/6, q = 5/6

–: End of Self Assessment and Revision on Theoritical Probability Distribution Class 12 OP Malhotra ISC Maths Solutions :–

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