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Two Dice Are Thrown Together and the Total Score is Noted. the Event E, F and G Are "A Total of 4", "A Total of 9 Or More", and "A Total Divisible by 5", Respectively. - Mathematics

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Question

Two dice are thrown together and the total score is noted. The event EF 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.   

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Solution

We have,

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

n(S) = 36

E = "a total of 4" = {(1, 3), (2, 2), (3, 1)} i.e. n(E) = 3
F = "a total of 9 or more = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)} i.e. n(F) = 10
G = "a total divisible by 5" = {(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)} i.e. n(G) = 7

Now,

\[P\left( E \right) = \frac{3}{36} = \frac{1}{12}, \]
\[P\left( F \right) = \frac{10}{36} = \frac{5}{18} \text{ and } \]
\[P\left( G \right) = \frac{7}{36}\]
\[\text{ Also } , \]
\[E \cap F = \phi, E \cap G = \phi \text{ and }  F \cap G = \left\{ \left( 4, 6 \right), \left( 5, 5 \right), \left( 6, 4 \right) \right\} \text{ i . e . n} \left( F \cap G \right) = 3\]
\[\text{ Since } , P\left( F \right) \times P\left( G \right) = \frac{5}{18} \times \frac{7}{36} = \frac{35}{648} \neq \frac{3}{36}\]
\[i . e . P\left( F \right) \times P\left( G \right) \neq P\left( F \cap G \right)\]
\[\text{ So, no pair is independent } .\]

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Problems based on Probability
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Chapter 31: Probability - Exercise 31.4 [Page 55]

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RD Sharma Mathematics [English] Class 12
Chapter 31 Probability
Exercise 31.4 | Q 24 | Page 55
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