Answer: 0.08192
Step-by-step explanation:
Given: Probability of free throws = 0.80
Then, probability of missing the free throw = 1-0.80=0.20
Now, According to the negative binomial probability distribution, the probability that the rth success is achieved in xth trial is given by ;-
[tex]P(X=x)= pC_{r-1}^{x-1}p^{r-1}(1-p)^{(x-r}[/tex]
Put p = 0.20 , x=5 and r =1, we get
The probability of this player missing a free throw for the first time on the fifth attempt as [tex](0.20)\ ^{5-1}C_{1-1}(0.20)^{1-1}(1-0.20)^{5-1}[/tex]
[tex]= (0.20)\ ^{4}C_{0}(0.20)^{0}(0.80)^{4}\\\\=(0.20)(1)(0.80)^4\\\\=(0.20)(0.4096)\\\\=0.08192[/tex]
hence the required probability =0.08192
