Question

A softball pitcher has a 0.507 probability of throwing a strike for each pitch. If the softball pitcher throws 15 pitches, what is the probability that more than 8 of them are strikes? (Round your answer to 3 decimal places if necessary.)

Answer 1

Answer:

P(X > 8) = 0.323

Step-by-step explanation:

We are given that a softball pitcher has a 0.507 probability of throwing a strike for each pitch. Also, the softball pitcher throws 15 pitches.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 15 pitches

            r = number of success = more than 8

           p = probability of success which in our question is probability of    

                 throwing a strike for each pitch, i.e., 0.507

LET X = Number of strikes

So, it means X ~ $Binom(n=15, p=0.507)$

So, Probability that more than 8 of them are strikes = P(X > 8)

P(X > 8) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

= $\binom{15}{9}0.507^{9} (1-0.507)^{15-9} +\binom{15}{10}0.507^{10} (1-0.507)^{15-10}+\binom{15}{11}0.507^{11} (1-0.507)^{15-11}+\binom{15}{12}0.507^{12} (1-0.507)^{15-12}+\binom{15}{13}0.507^{13} (1-0.507)^{15-13}+\binom{15}{14}0.507^{14} (1-0.507)^{15-14}+\binom{15}{15}0.507^{15} (1-0.507)^{15-15}$= $5005 \times 0.507^{9} \times 0.493^{6} +3003 \times 0.507^{10} \times 0.493^{5} +1365 \times 0.507^{11} \times 0.493^{4} +455 \times 0.507^{12} \times 0.493^{3} +105 \times 0.507^{13} \times 0.493^{2} +15 \times 0.507^{14} \times 0.493^{1} +1 \times 0.507^{15} \times 0.493^{0}$

= 0.323

Therefore, probability that more than 8 of them are strikes is 0.323.