Question #327786

If we toss two balanced dice, and let A be the event that the sum of the face values


of two dice is 8, and B be the event that the face value of the first one is 3. Calculate


P(A/B).

1
Expert's answer
2022-04-13T08:13:39-0400

A={(2,6),(3,5),(4,4),(5,3),(6,2)},B={(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)},AB={(3,5)}both the events A and Bhappened simultaneously.A=\{(2,6),(3,5),(4,4),(5,3),(6,2)\},\\ B=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\},\\ A\cap B=\{(3,5)\}-\text{both the events A and B} \\\text{happened simultaneously.}

Let's use the conditional probability formula:

P(AB)=P(AB)P(B).P(A|B)=\cfrac{P(A\cap B)}{P(B)}.

Total number of outcomes for a toss of two dice is n=66=36.n=6\cdot6=36.

So,

P(AB)=136,P(B)=636,P(AB)=1/366/36=16.P(A\cap B)=\cfrac{1}{36},\\ P(B)=\cfrac{6}{36},\\ P(A|B)=\cfrac{1/36}{6/36}=\cfrac{1}{6}.


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