Question

Winston (a dog) loves to play fetch with his ball! Sometimes Winston can even catch the ball in the air with probability 0.35.

You throw Winston the ball four times. Let X = the number of mid-air catches that Winston makes over four attempts.

a) Explain why X is a binomial random variable.

b) Compute the probability that Winston catches the ball in the air at least twice (i.e. P(X>1)).

c) On average, how many catches do you expect Winston to make (i.e. E(X))?

Answer 1

Solution(a)

X is a bionomial random variable because it is satisfying below conditions:

1. No.of observations is fixed I.e. 4

2. Probability for each and every trail is same I.e. 0.35

3. Each trail has one of two outcomes I.e. success or failure

4. Each observation is independent.

So Variable X meeting all conditions so this is a Bionomial random variable

Solutuin(b)

P= 0.35 and q = 0.65

P(X>1)= 1-p(X=0) +P(X=1)

=1 - 4C0 *(0.35)^0 *(0.65)^4 - 4C1*(0.35)^1 *(0.65)^3

1- (1*1*0.1785)-(4*0.35*0.2746)

=1-0.1785-0.384475 =0.4370

So there is 43.70% probability that Winston catches the ball in the air at least twice.

Solution(C)

On average, catches do you expect Winston to make = (no. Of sample * Probability) =(4*0.35) =1.40