(p + q)^5 = p^5 + (5p^4)(q) + (10p^3)(q^2) + (10 p^2)(q^3) + 5p(q^4) + q^5
If p = 0.3, then q = 1 - p = 1 - 0.3 = 0.7.
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1. Let X denote the random variable that represents the number of successes.
2. Let us have values of X:
ranging from no success X=0 to 5 successes X = 5.
3. Calculate each probability by mutlplying the probability of success ( in this case p = .3) for each success times the probability of failure (1-.3 = .7) for each failure.
4. Then, let us include all the rearrangements for our particular value of X.
For X= 0, there is no success, so the calculation is .7^5.
For X=1, we can have 1 success and 4 failures. .3^1 * .7^4, but we can rearrange the successes 5 ways too,
So, the P(X=1) is 5 * .3^1 * .7^4.
5. Work out for the rearrangements:
Use the combination rule nCr for each value.
Then the six calculations are
.7^5 = .16807
5C1 .3^1 .7^4
5C2 .3^2 .7^3
5C3 .3^3 .7^2
5C4 .3^4 .7^1
5C5 .3^5
6. Using a Ti-84 to do the actual calculations binompdf (5,.3,X) with X ranging from 0 to 5
It makes up:
.16807 .36015 .3087 .1323 .02835 .00243
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