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Binomial expansion

Could you guys please help me with the second part of this question.

The questions state as follows:

4.
A.

fully expand (p+q)^5

I have done this, using the binomial expansion theorem and have gotten an answer of:

p^5 +5p^4q + 10p^3q^2 + 10p^2q^3 + 5pq^4 + q^5

B.

a fair 4 sided die, numbered 1, 2, 3, and 4, is rolled 5 times. Let p represent the probability that the number 4 is rolled on a given roll, and let q represent the probability that the number 4 is not rolled on a given roll.

Using the first 3 terms of the binomial expansion from part a, find the probability that the number 4 is rolled at least 3 times.

How do I answer part B?

Thank you

Reply 1

old_engineer

p^5 is the probability of five successive 4's

5p^4q is the probability of of four 4's and one "no 4" out of five rolls, taking account of the five ways this can happen

etc

"at least three times" means three times or four times or five times.

Does that help?

Reply 2

YourGoddamnRight

Original post by old_engineer

does this mean that the probability for the first 3 terms is 12/15 ?

Reply 3

old_engineer

Original post by YourGoddamnRight

does this mean that the probability for the first 3 terms is 12/15 ?

No, that's not what I get. Can I suggest that you post your working, or perhaps state what you get for the three terms separately? Also perhaps state the values you're using for p and q.

(edited 6 years ago)

Reply 4

YourGoddamnRight

Original post by old_engineer

No, that's not what I get. Can I suggest that you post your working, or perhaps state what you get for the three terms separately?

sorry, I added the denominator of each, idk why.

but I got the answer of 12/5 from:

P^5 = probability of number 4 = 5/5
5p^4q = probability of number 4 = 4/5
10p^3q^2 = probability of number 4 = 3/5

5/5 + 4/5 + 3/5 = 12/5

Reply 5

old_engineer

Original post by YourGoddamnRight

Err no. The probability of anything happening can't be greater than 1. To set you going in the right direction, p = (1/4). It's a fair four-sided die so the probability of any particular value (e.g. 4) coming up in one throw must be (1/4). The probability of five successive 4's is then p^5 which is equal to (1/4)^5. Can you take it from there?

Reply 6

YourGoddamnRight

Original post by old_engineer

so p = 1/4 and q = 3/4.

so do I just substitute those values for the first 3 terms?

which would give me

(1/4)^5 + 5(1/4)^4 (3/4) + 10(1/4)^3 (3/4)^2

this would give me

1/1024 + 15/1024 + 45/512

what would I do after this?

Reply 7

thotproduct

My Stats is a bit rusty but.
It's part of the Binomial Distribution.

Let X be the discrete random variable 'the number of 4s obtained when rolling the dice'

Think of it as X~B(n,p), where n is your number of trials and p is the probability of a success each individual time. If you have a high-end calculator, this can be done for you by inputting values and giving you a table.

In this instance, your n is 5 because there are 5 trials, p assumably is 1/4

So we have X~B(5,0.25)

If we're wanting P(X greater or equal to 3), we can input values and do 1-P(X lessthanequal to 2) for our probability.

Using your expansion, you only need to look at the terms here, if we're looking for 'at least 3 times', we're wanting these terms
p^5 +5p^4q + 10p^3q^2
i.e 'Probability you roll 5 4s, probability you roll four 4s and 1 of something else, probability you roll 3 4s and 2 of something else'.

Alternatively, you could do 1 minus the last 3 terms
10p^2q^3 + 5pq^4 + q^5
i.e the probability you roll 2 4s and 3 of something else, 1 4 and 4 of something else, or 5 of something else, i.e your failing terms, by subtracting that from 1 you will get your success terms.

Reply 8

old_engineer

Original post by YourGoddamnRight

That's looking much better. To finish off, you just need to express the third of your terms with a denominator of 1024 rather than 512, then you can simply add the three terms together to give n/1024, which is the required answer.

Reply 9

YourGoddamnRight

Original post by old_engineer

Thank you so much.

Reply 10

the bear

Original post by YourGoddamnRight

1/1024 + 15/1024 + 45/512

what would I do after this?

that is your answer...

Reply 11

Elthera1

Hey! Can I ask you where did you get this exercise from?

Reply 12

sumayaxxxx

Original post by YourGoddamnRight

hey i was wondering if u have the differentiation version of this.

Reply 13

c00li9

Original post by YourGoddamnRight

Hello, I dont understand why substituting those values would give us the probability of there being a 4 rolled at least 3 times. Will someone explain please?

Reply 14

old_engineer

Original post by c00li9

Hello, I dont understand why substituting those values would give us the probability of there being a 4 rolled at least 3 times. Will someone explain please?

The first three terms of the binomial expansion give us the probabilities that there will be five, four, and three "4"s thrown respectively. Together, they give us the probability that at least three "4"s will be thrown.

This thread is three years old, by the way.