Binomial Probabilities?

Question

I need to find the indicated binomial probability? Help please

A surgical technique is performed on 7 patients. You are told there is a 70% chance of success. Find the probability that the surgery is successful for

A.) Exactly 5 patients
B.) Atleast 5 patients
C.) Less than 5 patients

Answer 1

The way to deal with the binomial theorem is to expand the term (p + q)^n, there p = probability of success, q = probability of failure = 1 - p, and n is the number of trials. This leads to the summation expression:

SIGMA (i=0 to n) (p^i * q^(n-i)) * n!/(i! * (n-i)!

The probability of success of m is found from the monomial that the term p^m. This is in general so far.

For your question, p = 0.70, q = 0.30, and n = 7. For (a), you are solving for the term containing p^5. For (b), it will be the sum of the terms containing p^5, p^6, and p^7. For (c), it will be the sum of the other five terms, but you can subtract (b)'s result from 1 to get the probability.

Hope that helps.

Answer 2

7 patients ... binomial ....so the pascal's line looks like this ....
1 7 21 35 35 21 7 1 ( 0,1,2,3,4,5,6,7)
so...
A) exactly 5 patients = 21 * (.7)^5 (.3)^2 (21 combo's of 5 successes & 2 failures)
21*.7*.7*.7*.7*.7 *.3*.3 = .3176532

B) at least 5 .... = P(5)+P(6)+P(7)
p(5) = .3176532
p(6) = 7*.7*.7*.7*.7*.7*.3 .2470629
P(7) = .7*.7*.7*.7*.7*.7*.7 .0823543
__________
0.6470704

C) less than 5 ... = 1 - (at least 5)
1 - .6470704 = 0.3529296

Answer 3

C