# Probability - Coins

• Dec 25th 2006, 08:43 AM
WeeG
Probability - Coins
I need a wee help with this little thing:

A fair coin is being thrown until H comes up for the first time. Let N be the number of times needed for H to turn up. If N=n, we throw n fair dices. Let S be the sum of the numbers turning up on the N dices.

a. find the probability that N=1 if you know that S=4, and the probability that S=6 if you know that N is an even number.

b.Show that the probability that the biggest number the appears on one of the dices is smaller or equal to K is K/12-k ( K is between 1 to 6 )

c.Find the probability that the biggest number that apears on at least one of the dices is K.

I have tried to solve "a" with Bayes's Theorem, but I don't know how to calculate P(S=4)....

cheers
( and sorry about my english...)
• Jan 11th 2007, 02:11 PM
F.A.P
Quote:

Originally Posted by WeeG
I need a wee help with this little thing:

A fair coin is being thrown until H comes up for the first time. Let N be the number of times needed for H to turn up. If N=n, we throw n fair dices. Let S be the sum of the numbers turning up on the N dices.

a. find the probability that N=1 if you know that S=4, and the probability that S=6 if you know that N is an even number.

Use Baye's Theorem
$P(N=1|S=4)=\frac{P(S=4|N=1)P(N=1)}{P(S=4)}$, condition on N in the denominator to get

$P(N=1|S=4)=\frac{P(S=4|N=1)P(N=1)}{\sum_{k=1}^{\in fty}P(S=4|N=k)P(N=k)}$

Now P(S=4|N=k) is non-zero only for N = 1, 2, 3, 4. Thus the denominator d becomes
$d={\sum_{k=1}^{4}P(S=4|N=k)P(N=k)}=\frac{1}{6}\fra c{1}{2}+\frac{3}{6^2}\frac{1}{4}+\frac{3}{6^3}\fra c{1}{8}+\frac{1}{6^4}\frac{1}{16}\approx0.106$

Finally $P(N=1|S=4)=\frac{1/6\cdot1/2}{d}=\frac{1728}{2197}\approx0.787$

For the next part we know that N must be 2, 4 or 6.

$P(S=6|N even)=\sum_{k=2,4,6}^{}P(S=6|N=k)P(N=k)$
$P(S=6|N even)=\frac{5}{6^2}\frac{1}{4}+\frac{10}{6^4}\frac {1}{16}+\frac{1}{6^6}\frac{1}{64}\approx0.789$
• Jan 16th 2007, 12:57 PM
WeeG
cheers
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