A pair of dice is thrown. Find the probability of getting 7 as the sum

1.	A pair of dice is thrown. Find the probability of getting 7 as the sum, if it is known that the second die always exhibits an odd number.
Answer» We know total possible outcomes when two dice are thrown = 36 A = sum is 7 {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1l} = 6 P(A) = \(\cfrac6{36}\) B =second die exhibit odd number {(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5),(4,1),(4,3),(4,5),(5,1),(5,3),(5,5),(6,1),(6,3),(6,5)} = 18 P(B) = \(\cfrac{18}{36}\) (A ∩ B) = sum of two number is 7 and also second die exhibit odd number = {(2,5),(4,3),(6,1)} = 3 P(A ∩ B) = \(\cfrac{3}{36}\) Therefore, \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\) = \(\cfrac{\frac3{36}}{\frac{18}{36}}\) = \(\cfrac{3}{18}=\cfrac16\)

A pair of dice is thrown. Find the probability of getting 7 as the sum, if it is known that the second die always exhibits an odd number.

Answer»

We know total possible outcomes when two dice are thrown = 36

A = sum is 7

{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1l} = 6

P(A) = \(\cfrac6{36}\)

B =second die exhibit odd number

{(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5),(4,1),(4,3),(4,5),(5,1),(5,3),(5,5),(6,1),(6,3),(6,5)} = 18

P(B) = \(\cfrac{18}{36}\)

(A ∩ B) = sum of two number is 7 and also second die exhibit odd number

= {(2,5),(4,3),(6,1)} = 3

P(A ∩ B) = \(\cfrac{3}{36}\)

Therefore, \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)

= \(\cfrac{\frac3{36}}{\frac{18}{36}}\) = \(\cfrac{3}{18}=\cfrac16\)

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