Posted by **Tom** on Wednesday, May 21, 2008 at 10:55pm.

Sorry for posting so many topics, but I really need help on this stuff. I recently lost my work I did on a seperate sheet from 2 weeks ago, and now I can't seem to find the answer again.

a fair six-sided die is tossed seven times in a row.

A) What is the probability that a "4" comes up exactly four times? (I got 4375/279936)

B) What is the probability that a "4" comes up an even number of times? (I got 2315/4374)

C) What is the probability that a number higher than "4" Comes up exactly four times (I got 280/2187)

Can anyone help me with this. I need to know how I got these answers again.

- Math -
**Writeacher**, Wednesday, May 21, 2008 at 10:57pm
Assistance needed.

- Test tomm. Need help *Urgent* -
**Reiny**, Thursday, May 22, 2008 at 8:38am
Tom, I answered part a) in

http://www.jiskha.com/display.cgi?id=1211429814

for b) using the same argument it would be

prob(exactly 2 times) + prob(exactly 4 times) + prob(exactly 6 times)

= C(7,2)(1/6)^2(5/6)^5 + C(7,4)(1/2)^4(5/6)^3 + C(7,6)(1/6)^6(5/6)

= 70035/279936 which is not what you had

but check my calculations.

- Test tomm. Need help *Urgent* -
**drwls**, Thursday, May 22, 2008 at 9:22am
B) I believe the middle term in Reiny's formula should be

C(7,4)(1/6)^4(5/6)^3

= C(7,4)*5^3/6^7

= 35*125/279,936 = 4375/279,936

which is the same as the answer to (A)

C(7,2)(1/6)^2(5/6)^5 = 21*3125/279,936

= 65625/279,936

C(7,6)(1/6)^6(5/6) = 7*5/279,936

= 35/279,936

Total = 70035/279936 = 0.25018

I agree with Reiny's final (B) answer, so the (1/2)^4 was probably a typo that should have been (1/6)^4

C) Since is equally likely to get "5" or "6" four times as it is to get "4" four times, just double the answer to (A). That does not agree with your answer.

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