Class 12: Maths Chapter 7 solutions. Complete Class 12 Maths Chapter 7 Notes.
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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The Maharashtra State Board of Secondary and Higher Secondary Education or MSBSHSE (Marathi: महाराष्ट्र राज्य माध्यमिक आणि उच्च माध्यमिक शिक्षण मंडळ), is an autonomous and statutory body established in 1965. The board was amended in the year 1977 under the provisions of the Maharashtra Act No. 41 of 1965.
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