Question

A new surgery is successful 70% of the time , if the results of 8 such surgeries are randomly sampled, what is the probability that more than 5 of them are successful?

Answer 1

P (x>5) = 0.5324014=0.523

Explanation:

Here success = p= 0.7 and failure = q= 1-p= 1-0.7 = 0.3

Number of surgeries= n= 8

Applying the binomial distribution

( p+q) ^n= nC x p^x . q^n-x

More than 5 means 6,7, and 8

( p+q) ^8= 8C 5 (0.7)^5 . (0.3)^3

=56* 0.16807*0.027

= 0.254121

( p+q) ^8= 8C 6 (0.7)^6 . (0.3)^2

= 28* 0.1177* 0.09

= 0.296604

( p+q) ^8= 8C 7 (0.7)^7 . (0.3)^1

= 8* 0.0824* 0.3

= 0.19776

( p+q) ^8= 8C8 (0.7)^8 . (0.3)^0

= 1* 0.05765* 1

=0.05765

Adding the above all probabilities gives

P (x>5) = P (x=6) + P(x=7) + P(x=8)

= 0.296604+0.19776+0.05765

= 0.5324014