Answer:
The required probability is 0.0155.
Step-by-step explanation:
We are given that the one-year survival rate for pancreatic cancer is 20%.
Of 12 people diagnosed with pancreatic cancer at one hospital, 6 survived one year.
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 of trials (samples) taken = 12 people
r = number of success = 6 survived
p = probability of success which in our question is probability of
one-year survival rate for pancreatic cancer, i.e; p = 20%
Let X = Number of people survived one year
So, X ~ Binom(n = 12 , p = 0.20)
Now, the probability that 6 survived one year is given by = P(X = 6)
P(X = 6) = [tex]\binom{12}{6} \times 0.20^{6} \times (1-0.20)^{12-6}[/tex]
= [tex]924 \times 0.20^{6} \times 0.80^{6}[/tex]
= 0.0155
