A die is thrown twice and the sum of the numbers appearing is observed to be 6
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(1,5) (2,4) (3,3) (4,2) (5,1)
P(E)= no. Of favorable outcomes /no. of possible outcomes
F. O=5
P.O=6^2=36
P(E)=5/36
I hope this answer helps you buddy
P(E)= no. Of favorable outcomes /no. of possible outcomes
F. O=5
P.O=6^2=36
P(E)=5/36
I hope this answer helps you buddy
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