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An unbiased coin is tossed. If the result is head, a pair of unbiased dice are rolled and the number obtained by adding the number on the two faces are noted. If the result is a tail, a card from a well-shuffled pack of 11 cards numbered 2, 3, 4……, 12 is picked & the number on the card is noted. What is the probability that the number noted is 7 or 8.
Let us define the events A : head appears. B : Tail appears C : 7 or 8 is noted. We have to find the probability of C i.e. P(C) P(C) = P(A) P (C/A) + P(B) P(C/B) Now we calculate each of the constituents one by one P(A) = probability of appearing head =1/2 P(C/A) = Probability that event C takes place i.e. 7 or 8 being noted when head is already appeared. (If something is already happen then it becomes certain, i.e. now it is certain that head is appeared we have to certainly roll a pair of unbiased dice). =11/36 (since (6, 1) (1, 6) (5, 2) (2, 5) (3, 4) (4, 3) (6, 2) (2, 6) (3, 5) (5, 3) (4, 4) i.e. 11 favourable cases and of course 6 × 6 = 36 total number of cases) Similarly, P(B) = 1/2 P(B/C) = 2/11 (Two favourable cases (7 and 8) and 11 total number of cases). Hence, P(C) = 1/2 * 11/36 + 1/2 * 2/11 = 193/792
Let us define the events
A : head appears.
B : Tail appears
C : 7 or 8 is noted.
We have to find the probability of C i.e. P(C)
P(C) = P(A) P (C/A) + P(B) P(C/B)
Now we calculate each of the constituents one by one
P(A) = probability of appearing head =1/2
P(C/A) = Probability that event C takes place i.e. 7 or 8 being noted when head is already appeared. (If something is already happen then it becomes certain, i.e. now it is certain that head is appeared we have to certainly roll a pair of unbiased dice).
=11/36 (since (6, 1) (1, 6) (5, 2) (2, 5) (3, 4) (4, 3) (6, 2) (2, 6) (3, 5) (5, 3) (4, 4) i.e. 11 favourable cases and of course 6 × 6 = 36 total number of cases)
Similarly, P(B) = 1/2
P(B/C) = 2/11 (Two favourable cases (7 and 8) and 11 total number of cases).
Hence, P(C) = 1/2 * 11/36 + 1/2 * 2/11 = 193/792
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