The 3yearsurvival rate for a certain group of cancer patient

The 3-year-survival rate for a certain group of cancer patients is known to be 0.60. What is the probability (to four decimal places) that in a group of 10 patients, more than 5 patients survive at least 3 years (note: 3-year-survival is a binary outcome, and “success” is defined as survival).

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

P( X < = 5) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 10 5 ) * 0.6^5 * ( 1- 0.6 ) ^5 + ( 10 4 ) * 0.6^4 * ( 1- 0.6 ) ^6 + ( 10 3 ) * 0.6^3 * ( 1- 0.6 ) ^7 + ( 10 2 ) * 0.6^2 * ( 1- 0.6 ) ^8 + ( 10 1 ) * 0.6^1 * ( 1- 0.6 ) ^9 + ( 10 0 ) * 0.6^0 * ( 1- 0.6 ) ^10
= 0.367
P( X > 5) = 1 - P ( X <=5) = 1 -0.367 = 0.6331

P(more than 5 patients survive at least 3 years) = 0.6331