Binomial Probabilities? Help?

Question

Binomial Probabilities?
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

Ok, been awhile since I did one of these. But assuming 7 patients, there are 2,520 permutations, divided by 5! = 21 possible sets of 5 patients who could successfully have the operation.

Odds of the 5/2 success ratio = 0.0151263% * 21 = 31.76523%

I don't remember any practical way to do the rest of this and don't really have time at the moment, but I'm hoping this can at least get you started.

Answer 2

Ok, it's been a LONG time for me on this, so here goes. I start with the table (I'm doing it by hand, but could look it up)
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1

So that makes it
# success Factor
01
1 7
2 21
3 35
4 35
5 21 X (0.7)^5 x(0.3)^2 = 31.7652%
6 7 X (0.7)^6 x(0.3)^1 = 24.7063%
7 1 X (0.7)^6 x(0.3)^1 = 8.2354%

I filled in the formula for 5, 6, and 7 successes.


So for A) 5 successes is 21 X (0.7)^5 x(0.3)^2 = 31.7652% or 31.8%

For B) at least 5 patients is 64.707% or 64.7% (adding the three %'s up)

And C) less than 5 patients is 35.2931% or 35.3% (1 - 64.7%)

Cheers!