Answer:
The probability that Jason will get exactly 7 strikes out of 10 attempts is 0.117.
Step-by-step explanation:
We are given that Jason is a very good bowler and has proven over the course of a season of league play that he gets a STRIKE 50% of the time.
Also, Jason has been given 10 attempts.
The above situation can be represented through binomial distribution;
[tex]P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......[/tex]
where, n = number trials (samples) taken = 10 attempts
r = number of success = 7 strikes
p = probability of success which in our question is % of the time
he gets a strike, i.e; p = 50%
Let X = Number of strikes Jason get
So, X ~ Binom(n = 10, p = 0.50)
Now, probability that Jason will get exactly 7 strikes out of 10 attempts is given by = P(X = 7)
P(X = 7) = [tex]\binom{10}{7} \times 0.50^{7} \times (1-0.50)^{10-7}[/tex]
= [tex]120 \times 0.50^{7} \times 0.50^{3}[/tex]
= [tex]120 \times 0.50^{10}[/tex]
= 0.117
Therefore, the probability that Jason will get exactly 7 strikes out of 10 attempts is 0.117.
