I roll a pair of dice, What is the probability that
A. The sum of the pips on the top faces is 7 and double is rolled ?
I said P ( sum of 7 or double is rolled ) = 0
So when we try to compute using the inclusion/exclusion principle the last term of the formula will go away right ?
P (sum of 7 or doubles ) = P ( sum of 9 ) / P (doubles ) - P( sump of 7 or double )
Something devided by 36 + someting else divided by 36 as well minus 0 right ?
THE THING IS I DON'T KOW BY WHAT SHOULD I DIVIDED 36 ?
Thank you and take care !
+130555 I'm not sure as to what you're calculating, but, if we're asking the probability of a 7 or the probability of a double, we have :
P ( 7 or a double) = P(7) + P(double) - P(7 and a double)
The last term is 0.......no doubles equal 7
P(7) .......there are 36 possibilities when we roll 2 dice........the probability of a 7 = 6/36 ....and the probabilty of a double = 6/36
So
P(7 or a double) = 6/36 + 6/36 - 0 = 12/36 = 1/3
