**Chapter 7 Probability Distributions Miscellaneous Exercise 7**

## Chapter 7 Probability Distributions Miscellaneous Exercise 7

**(I) Choose the correct option from the given alternatives:**

**Question 1.P.d.f. of a c.r.v. X is f(x) = 6x(1 – x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere) If P(X < a) = P(X > a), then a =**

**Question 3.**

**Question 5.**

**Question 7.**

**Question 8.If the d.r.v. X has the following probability distribution:**

**Question 9.If the d.r.v. X has the following probability distribution:then k =**

**Question 10.Find the expected value of X for the following p.m.f.**

(a) 0.85

(b) -0.35

(c) 0.15

(d) -0.15

**Answer:**

(b) -0.35

**(II) Solve the following:**

**Question 1.Identify the random variable as either discrete or continuous in each of the following. If the random variable is discrete, list its possible values:(i) An economist is interested in the number of unemployed graduates in the town of population 1 lakh.(ii) Amount of syrup prescribed by a physician.(iii) The person on a high protein diet is interesting to gain weight in a week.(iv) 20 white rats are available for an experiment. Twelve rats are males. A scientist randomly selects 5 rats, the number of female rats selected on a specific day.(v) A highway-safety group is interested in studying the speed (in km/hr) of a car at a checkpoint.Solution:**

(i) Let X = number of unemployed graduates in a town.

Since the population of the town is 1 lakh, X takes the finite values.

∴ random variable X is discrete.

Range = {0, 1, 2, …, 99999, 100000}.

(ii) Let X = amount of syrup prescribed by a physician.

Then X takes uncountable infinite values.

∴ random variable X is continuous.

(iii) Let X = gain of weight in a week

Then X takes uncountable infinite values

∴ random variable X is continuous.

(iv) Let X = number of female rats selected on a specific day.

Since the total number of rats is 20 which includes 12 males and 8 females, X takes the finite values.

∴ random variable X is discrete.

Range = {0, 1, 2, 3, 4, 5}

(v) Let X = speed of .the car in km/hr.

Then X takes uncountable infinite values

∴ random variable X is continuous.

**Question 2.The probability distribution of discrete r.v. X is as follows:(i) Determine the value of k.(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).Solution:**

**Question 3.The following is the probability distribution of X:Find the probability that(i) X is positive(ii) X is non-negative(iii) X is odd(iv) X is even.Solution:**

(i) P(X is positive) = P(X = 1) + P(X = 2) + P(X = 3)

= 0.25 + 0.15 + 0.1

= 0.50

(ii) P(X is non-negative)

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 0.20 + 0.25 + 0.15 + 0.1

= 0.70

(iii) P(X is odd)

= P(X = -3) + P(X = -1) + P(X = 1) + P(X = 3)

= 0.05 + 0.15 + 0.25 + 0.1

= 0.55

(iv) P(X is even)

= P(X = -2) + P(X = 0) + P(X = 2)

= 0.10 + 0.20 + 0.15

= 0.45.

**Question 4.**

= P(X = 5) + P(X = 4) + P(X = 3)

= P(X ≥ 3)

∴ P(X ≤ 2) = P(X ≥ 3).

**Question 5.In the p.m.f. of r.v. XFind a and obtain c.d.f. of X.Solution:**

For p.m.f. of a r.v. X

∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1

Let F(x) be the c.d.f. of X.

Then F(x) = P(X ≤ x)

Hence, the c.d.f. of the random variable X is as follows:

**Question 6.A fair coin is tossed 4 times. Let X denote the number of heads obtained. Write down the probability distribution of X. Also, find the formula for p.m.f. of X.Solution:**

When a fair coin is tossed 4 times then the sample space is

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}

∴ n(S) = 16

X denotes the number of heads.

∴ X can take the value 0, 1, 2, 3, 4

When X = 0, then X = {TTTT}

∴ n (X) = 1

∴ the probability distribution of X is as follows:

Also, the formula for p.m.f. of X is

**Question 7.Find the probability distribution of the number of successes in two tosses of a die, where success is defined as(i) number greater than 4(ii) six appear on at least one die.Solution:**

When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.

Let X be the random variable, which represents the number of successes.

Here, success refers to the number greater than 4.

P(X = 0) = P(number less than or equal to 4 on both the tosses)

P(X = 1) = P(number less than or equal to 4 on first toss and greater than 4 on second toss) + P(number greater than 4 on first toss and less than or equal to 4 on second toss)

Thus, the probability distribution is as follows:

(ii) Here, success means six appears on at least one die.

**Question 8.A random variable X has the following probability distribution:Determine:**

(i) k

(ii) P(X > 6)

(iii) P(0 < X < 3).

**Question 9.The following is the c.d.f. of a r.v. X:Find(i) p.m.f. of X(ii) P( -1 ≤ X ≤ 2)(iii) P(X ≤ X > 0).Solution:**

(i) From the given table

F(-3) = 0.1, F(-2) = 0.3, F(-1) = 0.5

F(0) = 0.65, f(1) = 0.75, F(2) = 0.85

F(3) = 0.9, F(4) = 1

P(X = -3) = F(-3) = 0.1

P(X = -2) = F(-2) – F(-3) = 0.3 – 0.1 = 0.2

P(X = -1) = F(-1) – F(-2) = 0.5 – 0.3 = 0.2

P(X = 0) = F(0) – F(-1) = 0.65 – 0.5 = 0.15

P(X = 1) = F(1) – F(0) = 0.75 – 0.65 = 0.1

P(X = 2) = F(2) – F(1) = 0.85 – 0.75 = 0.1

P(X = 3) = F(3) – F(2) = 0.9 – 0.85 = 0.1

P(X = 4) = F(4) – F(3) = 1 – 0.9 = 0.1

∴ the p.m.f of X is as follows:

(ii) P(-1 ≤ X ≤ 2) = P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2)

= 0.2 + 0.15 + 0.1 + 0.1

= 0.55

(iii) (X ≤ 3) ∩ (X > 0)

= { -3, -2, -1, 0, 1, 2, 3} n {1, 2, 3, 4}

= {1, 2, 3}

**Question 10.Find the expected value, variance, and standard deviation of the random variable whose p.m.f’s are given below:Solution:**

(i) We construct the following table to find the expected value, variance, and standard deviation:

(ii) We construct the following table to find the expected value, variance, and standard deviation:

(iii) We construct the following table to find the expected value, variance, and standard deviation:

(iv) We construct the following table to find the expected value, variance, and standard deviation:

**Question 11.A player tosses two wins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears and ₹ 2 if no head appears. Find the expected winning amount and variance of the winning amount.Solution:**

When a coin is tossed twice, the sample space is

S = {HH, HT, TH, HH}

Let X denote the amount he wins.

Then X takes values 10, 5, 2.

We construct the following table to calculate the mean and the variance of X:

= 38.5 – 30.25

= 8.25

∴ Hence, expected winning amount = ₹ 5.5 and variance of winning amount = ₹ 8.25.

**Question 12.**

We construct the following table to calculate the mean and variance of X:

From the table

= 1 – 0

= 1

Hence, E(X) = 0, Var (X) = 1.

**Question 13.Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.f(x) = k(4 – ), -2 ≤ x ≤ 2 and = 0 otherwise.Compute(i) P(X > 0)(ii) P(-1 < X < 1)(iii) P(X < -0.5 or X > 0.5).Solution:**

(i) P(X > 0)

(ii) P(-1 < X < 1)

(iii) P(X < -0.5 or X > 0.5)

**Question 14.Solution:**

**Question 15.Solution:**

Since f is p.d.f. of the r.v. X,