Question

Can someone help me with this question?

Answer 1

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;

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

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)  =  $\binom{12}{6} \times 0.20^{6} \times (1-0.20)^{12-6}$

                             =  $924 \times 0.20^{6} \times 0.80^{6}$

                             =  0.0155