# Dice and the binomial distribution.

• May 12th 2010, 02:08 PM
Zoheb Imran
Dice and the binomial distribution.
A Die is Biased so that the probability of throwing a 5 is 0.75 and the probabilities of throwing a 1,2,3,4 or 6 are all equal.

i. The Die is thrown three times. Find the probability that the result is a 1 followed by a 5 followed by any even number.
My Attempt: P of 5: 0.75 for 1,2,3,4,6: 0.05
So P( 1 followed by 5 and then even)= 0.05 x .075 x (0.05+0.05+0.05) (adding because of "or")
= 0.0005625

ii. Find the Probability that, out of 10 throws of this die, at least 8 result in a 5.
My attempt: using binomial p = .75 q=.25
n=10
p(x ≥ 8) = p(x=8) + p(x=9) + p(x=10)
= 10C8 (.75)^8 (0.25)^2 + 10C9 (0.75)^9 (0.25)+ 10C10 (0.75)^10 (0.25)^0

iii. The die is thrown 90 times. Using an appropriate approximation, find the probability that a 5 is thrown more than 60 times.
Please solve this one. We have studied binomial and normal Distribution Only :)

Please check my attempts and tell me is it correct? if wrong tell me where did i go wrong. and please give ur advices too. My Exam is on 24th May So need Tips
• May 12th 2010, 02:52 PM
pickslides
Quote:

Originally Posted by Zoheb Imran

iii. The die is thrown 90 times. Using an appropriate approximation, find the probability that a 5 is thrown more than 60 times.
Please solve this one. We have studied binomial and normal Distribution Only :)

You can use a normal approximation to the binomial

In this case $n= 90$ and $p=0.75$

$
P(X>60) = P\left(Z>\frac{60-\mu}{\sigma}\right)\approx P\left(Z\geq\frac{(60+0.5)-np}{\sqrt{np(1-p)}}\right)=\dots

$
• May 12th 2010, 11:52 PM
downthesun01
Yes, the first two are correct and pickslides provided the steps to complete part 3 using normal approximation of the binomial distribution.
• May 13th 2010, 06:04 PM
matheagle
Quote:

Originally Posted by pickslides
You can use a normal approximation to the binomial

In this case $n= 90$ and $p=0.75$

$
P(X>60) = P\left(Z>\frac{60-\mu}{\sigma}\right)\approx P\left(Z\geq\frac{(60+0.5)-np}{\sqrt{np(1-p)}}\right)=\dots

$

correct but the approximation appears when you replace the binomial with the normal...

$
P(Binomial >60) \approx P(Normal>60.5)=P\left(Z>\frac{(60+0.5)-np}{\sqrt{np(1-p)}}\right)$