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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) and decide which pairs of events, if any, are independen

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.

Answers (1)

Given-

Two dice are drawn together i.e. n(S)= 36

S is the sample space

E = a of total 4

F= a total of 9 or more

G= a total divisible by 5

Therefore, for E,

E = a of total 4

∴E = {(2,2), (3,1), (1,3)}


∴n(E) = 3

For F,

F= a total of 9 or more

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

∴n(F)=10

For G,

G = a total divisible by 5

∴ G = {(1,4), (4,1), (2,3), (3,2), (4,6), (6,4), (5,5)}
∴ n(G) = 7

Here, (E \cap F) = φ AND (E \cap G) = φ

Also, (F \cap G) = {(4,6), (6,4), (5,5)}
 \Rightarrow n (F \cap G) = 3 and (E \cap F \cap G) = φ

\\ P(E)=\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12} \\ P(F)=\frac{n(F)}{n(S)}=\frac{10}{36}=\frac{5}{18} \\ P(G)=\frac{n(G)}{n(S)}=\frac{7}{36} \\ P(F \cap G)=\frac{3}{36}=\frac{1}{12} \\ P(F) \cdot P(G)=\frac{5}{18} \times \frac{7}{36}=\frac{35}{648}

Therefore,

P (F \cap G) ≠ P(F). P(G)

Hence, there is no pair which is independent

Posted by

infoexpert22

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