Question

In a basketball match, a shooting guard can either throw the ball in the basket and score points for his team or miss it and not score. Each successful basket is worth one point. random variable, because the outcomes are The shooting guard's ball throw is an example of a Bernoulli binary Suppose the shooting guard successfully throws the ball in the basket 3 out of 10 times. Complete the probability distribution of the shooting guard's ball throws (X). Х Successful throw Unsuccessful throw P(X) The expected value of the shooting guard's throw is and its variance is

Answer 1

X	successful throw	unsuccessful throw
P(X)	0.3	0.7

Probsbility of successful throw

=P = successful throw / total Throw = 0.3

Probability of unsuccessful throw

= 1 - Probability of successful throw

= 0.7

2.

For a single trial, shooting guards throw is a bernoulli distribution, but for n trials it is sume of bernoulli distribution which is binomial distribution.

Expected value of binomial distribution = n*p

Expected value of shooting guards throw

= E(X) = n*p = 10 * 0.3 = 3

Variance of binomial distribution = n*p*(1-p)

Var(x) = 10 * 0.3 * 0.7 = 2.1

3.

Now there are two points for each throw

Hence we find expectation of 2*X

Hence,

Expected value of shooting guards throw

= E(2*X) = 2*E(X) = 2*3 = 6

Variance is

Var(2X) = 4*var(x) = 4 * 2.1 = 8.4

{since var(a*X) = a^2*var(x),

Where a is constant

E(a*X) = a*E(X)}