Question

Let X denote a random variable which corresponds to the number of players in a team who get injured during a game. If we assume that each player hasequal probability of being injured, which is 0.2 , and there are 10 of themparticipating in each game.
a. Is X a discrete or a continuous random variable?
b. What kind of distribution X follows?
c. Compute the probability that nobody or only one person will be injured in a game.
d. Write down the expression which gives the average value of X (also known as the expected value), what is the value of this expression?
e. Write down the expression that gives the variance of X (average squared distance from the mean ) what number is this expression equal to?
f. Assume that players can be ranked according to their performances and no two people are equally good. If two players get injured what is the probability of best player being one of them ? Note that all combinations of two people are equally likely to get injured.

Answer 1

Answer:

(a)

The values of X can be 0, 1, 2 , ..., 10 . So, X is a discrete random variable.

(b)

The distribution of X is Binomial distribution with the parameters n = 10 and p = 0.2

(c)

Probability that no one or one person will be injured = P(X = 0) + P(X = 1)

= 10C0 * 0.20 * (1 - 0.2)10-0 + 10C1 * 0.21 * (1 - 0.2)10-1

= 0.810 + 10 * 0.2 * 0.89

= 0.3758096

(d)

Average value of X = np

Average value of X = 10 * 0.2 = 2

(e)

Variance of X = np(1-p)

Variance of X = 10 * 0.2 * (1 - 0.2) = 1.6

(f)

Number of ways in which 2 people gets injured = 10C2 = 10! / ((10-2)! 2!) = (10 * 9) / (2 * 1) = 45

Assume the best player got injures, number of ways in which one people out of remaining 9 people gets injured = 9C1

= 9! / ((9-1)! 1!)

= 9

Probability that the best player got injured = Number of ways in which 1 people gets out of 9 and best person gets injured / Number of ways in which 2 people gets injured

= 9 / 45

= 0.2