# Binomial expansion

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Could you guys please help me with the

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

How do I answer part B?

Thank you

**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

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#2

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?

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?

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(Original post by

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?

**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?

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#4

(Original post by

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

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

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(Original post by

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?

**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?

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

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#6

(Original post by

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

**YourGoddamnRight**)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

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(Original post by

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?

**old_engineer**)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?

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?

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#8

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.

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.

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#9

(Original post by

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?

**YourGoddamnRight**)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?

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(Original post by

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.

**old_engineer**)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.

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#13

(Original post by

Could you guys please help me with the

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

How do I answer part B?

Thank you

**YourGoddamnRight**)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

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#14

**YourGoddamnRight**)

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?

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#15

(Original post by

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?

**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?

*at least*three "4"s will be thrown.

This thread is three years old, by the way.

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