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