YourGoddamnRight
Badges: 15
Rep:
?
#1
Report Thread starter 3 years ago
#1
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
2
reply
old_engineer
Badges: 11
Rep:
?
#2
Report 3 years ago
#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?
0
reply
YourGoddamnRight
Badges: 15
Rep:
?
#3
Report Thread starter 3 years ago
#3
(Original post by 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?
does this mean that the probability for the first 3 terms is 12/15 ?
0
reply
old_engineer
Badges: 11
Rep:
?
#4
Report 3 years ago
#4
(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.
0
reply
YourGoddamnRight
Badges: 15
Rep:
?
#5
Report Thread starter 3 years ago
#5
(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
0
reply
old_engineer
Badges: 11
Rep:
?
#6
Report 3 years ago
#6
(Original post by 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
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?
0
reply
YourGoddamnRight
Badges: 15
Rep:
?
#7
Report Thread starter 3 years ago
#7
(Original post by 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 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?
0
reply
thotproduct
Badges: 19
Rep:
?
#8
Report 3 years ago
#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.
1
reply
old_engineer
Badges: 11
Rep:
?
#9
Report 3 years ago
#9
(Original post by 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?
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.
1
reply
YourGoddamnRight
Badges: 15
Rep:
?
#10
Report Thread starter 3 years ago
#10
(Original post by 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.
Thank you so much.
0
reply
the bear
Badges: 20
Rep:
?
#11
Report 3 years ago
#11
(Original post by YourGoddamnRight)
1/1024 + 15/1024 + 45/512

what would I do after this?
that is your answer...
1
reply
Elthera1
Badges: 3
Rep:
?
#12
Report 3 years ago
#12
Hey! Can I ask you where did you get this exercise from?
0
reply
sumayaxxxx
Badges: 3
Rep:
?
#13
Report 2 years ago
#13
(Original post by 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
hey i was wondering if u have the differentiation version of this.
0
reply
c00li9
Badges: 3
Rep:
?
#14
Report 2 months ago
#14
(Original post by 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?
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?
0
reply
old_engineer
Badges: 11
Rep:
?
#15
Report 2 months ago
#15
(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.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Which of these would you use to help with making uni decisions?

Webinars (32)
13.73%
Virtual campus tours/open days (51)
21.89%
Live streaming events (22)
9.44%
Online AMAs/guest lectures (23)
9.87%
A uni comparison tool (55)
23.61%
An in-person event when available (50)
21.46%

Watched Threads

View All