Question

A binomial experiment consists of four independent trials. The probability of success in each trial is 9⁄25 . Find the probabilities of obtaining exactly 0 successes, 1 success, 2 successes, 3 successes, and 4 successes, respectively, in this experiment.
a) [0.1678, 0.0944, 0.3716, 0.0299, 0.0168]
b) [0.1678, 0.0105, 0.1593, 0.1194, 0.0168]
c) [0.1678, 0.3775, 0.3185, 0.1194, 0.0168]
d) [0, 0.1678, 0.3775, 0.3185, 0.1194]
e) [0.1678, 0.0944, 0.3185, 0.3318, 0.0168]
f) None of the above.

Answer 1

please help me...

Answer 2

The probability of obtaining k successes in n trials with probability of success p in each trial is
given by the formula
\[
\binom{n}{k} p^k (1-p)^{n-k}=\frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}
\]

Answer 3

Here n=4, p=9/25, k=0,1,2,3,4
You have all what you need to compute 5 numbers, for k=0,1,2,3,4
Compute them and see which of your five results match one of the choices

Answer 4

For k=0;
\[
\frac {4!}{0!(4-0)!}(9/25)^0(1-9/25)^4=(9/25)^0(16/25)^4=0.167772
\]

Answer 5

For k=1; you get
\[
4 (9/25)^1 (16/25)^3=0.377487
\]

Answer 6

Do k=3, 4, and 5 yourself and you will be done

Answer 7

Of course, you can use your calculator

Answer 8

I meant k=2,3 and 4 above