# Thread: Dice and the binomial distribution.

1. ## 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

2. 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

$

3. Yes, the first two are correct and pickslides provided the steps to complete part 3 using normal approximation of the binomial distribution.

4. 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)$