Class 12: Maths Chapter 7 solutions. Complete Class 12 Maths Chapter 7 Notes.
Contents
Maharashtra Board Solutions Class 12-Arts & Science Maths (Part 2): Chapter 7- Probability Distributions
Maharashtra Board 12th Maths Chapter 7, Class 12 Maths Chapter 7 solutions
Ex 7.1
Question 1.
Let X represent the difference between a number of heads and the number of tails when a coin is tossed 6 times. What are the possible values of X?
Solution:
When a coin is tossed 6 times, the number of heads can be 0, 1, 2, 3, 4, 5, 6.
The corresponding number of tails will be 6, 5, 4, 3, 2, 1, 0.
∴ X can take values 0 – 6, 1 – 5, 2 – 4, 3 – 3, 4 – 2, 5 – 1, 6 – 0
i.e. -6, -4, -2, 0, 2, 4, 6.
∴ X = {-6, -4, -2, 0, 2, 4, 6}.
Question 2.
An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Solution:
The urn contains 5 red and 2 black balls.
If two balls are drawn from the urn, it contains either 0 or 1 or 2 black balls.
X can take values 0, 1, 2.
∴ X = {0, 1, 2}.
Question 3.
State which of the following are not the probability mass function of a random variable. Give reasons for your answer.
Solution:
P.m.f. of random variable should satisfy the following conditions:
(a) 0 ≤ pi ≤ 1
(b) Σpi = 1.
(i)
(a) Here 0 ≤ pi ≤ 1
(b) Σpi = 0.4 + 0.4 + 0.2 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.
(ii)
P(X = 3) = -0.1, i.e. Pi < 0 which does not satisfy 0 ≤ Pi ≤ 1
Hence, P(X) cannot be regarded as p.m.f. of the random variable X.
(iii)
(a) Here 0 ≤ pi ≤ 1
(b) ∑pi = 0.1 + 0.6 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.
(iv)
Here ∑pi = 0.3 + 0.2 + 0.4 + 0 + 0.05 = 0.95 ≠ 1
Hence, P(Z) cannot be regarded as p.m.f. of the random variable Z.
(v)
Here ∑pi = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence, P(Y) cannot be regarded as p.m.f. of the random variable Y.
(vi)
(a) Here 0 ≤ pi ≤ 1
(b) ∑pi = 0.3 + 0.4 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.
Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.
Solution:
(i) For two tosses of a coin the sample space is {HH, HT, TH, TT}
Let X denote the number of heads in two tosses of a coin.
Then X can take values 0, 1, 2.
Ex 7.2
Question 1.
Verify which of the following is p.d.f. of r.v. X:
(i) f(x) = sin x, for 0 ≤ x ≤ π2
(ii) f(x) = x, for 0 ≤ x ≤ 1 and -2 – x for 1 < x < 2
(iii) fix) = 2, for 0 ≤ x ≤ 1.
Solution:
f(x) is the p.d.f. of r.v. X if
(a) f(x) ≥ 0 for all x ∈ R and
Hence, f(x) is the p.d.f. of X.
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Maharashtra Board Solutions Class 12-Arts & Science Maths (Part 2): Chapter 1- Differentiation
Chapterwise Maharashtra Board Solutions Class 12 Arts & Science Maths (Part 2) :
- Chapter 1- Differentiation
- Chapter 2- Applications of Derivatives
- Chapter 3- Indefinite Integration
- Chapter 4- Definite Integration
- Chapter 5- Application of Definite Integration
- Chapter 6- Differential Equations
- Chapter 7- Probability Distributions
- Chapter 8- Binomial Distribution
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