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Two dice are thrown together and the total score is noted. The events E, F and G are ‘a total of 4’, ‘a total of 9 or more’, and ‘a total divisible by 5’, respectively. Calculate P(E), P(F) and P(G) - Mathematics

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Question

Two dice are thrown together and the total score is noted. The events E, F and G are ‘a total of 4’, ‘a total of 9 or more’, and ‘a total divisible by 5’, respectively. Calculate P(E), P(F) and P(G) and decide which pairs of events, if any, are independent.

Sum
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Solution

Two dice are thrown together

∴ n(S) = 36

E = A total of 4 = {(2, 2), (1, 3), (3, 1)}

∴ n(E) = 3

F = A total of 9 or more

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

∴ n(F) = 10

G = A total divisible by 5

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

∴ n(G) = 7

Here, we see that (E ∩ F) = Φ and (E ∩ G) = Φ

And (F ∩ G) = {(4, 6), (6, 4), (5, 5)}

∴ n(F ∩ G) = 3 and (E ∩ F ∩ G) = Φ

∴ P(E) = `("n"("E"))/("n"("S")) = 3/36 = 1/12`

P(F) = `("n"("F"))/("n"("S")) = 10/36 = 5/18`

P(G) = `("n"("G"))/("n"("S")) = 7/36`

P(F ∩ G) = `3/36 = 1/12`

And P(F) . P(G) = `5/18 * 7/36 = 35/648`

Since, P(F ∩ G) ≠ P(F) . P(G)

Hence, there is no pair of independent events.

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Chapter 13: Probability - Exercise [Page 272]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 13 Probability
Exercise | Q 5 | Page 272
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