Question

The probability that a basketball athlete hit the rim of basketball is 0.9. Evaluate the probability distribution of the number of hitting times X if he shoots the basket twice independently.

Answer 1

Answer:

Probability of hitting two times is 0.81, one time is 0.18, zero times 0.01.

Step-by-step explanation:

We need to consider the even as binomial distribution function. Therefore, the probability distribution of the number of hitting times X are:

$P(X=0)=combination(2,0)*0.9^0*0.1^2=0.01\\P(X=1)=combination(2,1)*0.9^1*0.1^1=0.18\\P(X=2)=combination(2,2)*0.9^2*0.1^0=0.81$