In a single throw of two dice, determine the probability of getting a

1.	In a single throw of two dice, determine the probability of getting a total of 7 or 9.
Answer» Let S be the sample space. Then, n(S) = 6² = 36.Let ‘A’ and ‘B’ be the events of getting a total of 7 and 9 respectively. ∴ A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} ⇒ n(A) = 6 B = {(3, 6) (4, 5), (5, 4), (6, 3)}, ⇒ n(B) = 4 A ∩ B = ϕ ⇒ n(A ∩ B) = 0 ∴ P(A) = \(\frac{n(A)}{n(S)} = \frac{6}{36}, P(B) = \frac{n(B)}{n(S)} = \frac{4}{36}\) and O(A ∩ B) = \(\frac{n(A∩B)}{n(S)}= 0\) We know– P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = \(\frac{6}{36}+\frac{4}{36}-0\) = \(\frac{10}{36}\) = \(\frac{5}{18}\) ∴ The probability of getting a total of 7 or 9 is \(\frac{5}{15}\) .

In a single throw of two dice, determine the probability of getting a total of 7 or 9.

Answer»

Let S be the sample space. Then, n(S) = 6² = 36.Let ‘A’ and ‘B’ be the events of getting a total of 7 and 9 respectively.

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

⇒ n(A) = 6

B = {(3, 6) (4, 5), (5, 4), (6, 3)}, ⇒ n(B) = 4

A ∩ B = ϕ ⇒ n(A ∩ B) = 0

∴ P(A) = \(\frac{n(A)}{n(S)} = \frac{6}{36}, P(B) = \frac{n(B)}{n(S)} = \frac{4}{36}\)
and O(A ∩ B) = \(\frac{n(A∩B)}{n(S)}= 0\)

We know–

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{6}{36}+\frac{4}{36}-0\)

= \(\frac{10}{36}\)

= \(\frac{5}{18}\)

∴ The probability of getting a total of 7 or 9 is \(\frac{5}{15}\) .

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