Latest hpe Aptitude Question SOLUTION: an unbiased coin is tossed. if the result is a head, a pair o unbiased dice is rolled and the number of obtained is adding the numbers on two faces is noted if the result is a tail

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(#M40155988) HPE General Ability question probanility Keep an EYE

an unbiased coin is tossed. if the result is a head, a pair o unbiased dice is rolled and the number of obtained is adding the numbers on two faces is noted if the result is a tail, a card from a shuffled pack of a eleven cards 2, 3, 4, ......, 12 is picked and the number on the card is noted. the probability that the noted number is either 7 or 8 is

Asked In hpe sunitha (8 years ago)

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dhananjay kumar  (8 years ago)

find the answer to your question



E1 ≡ number noted is 7, E2 ≡ number notes is 8,


H ≡ getting head on coin, T ≡ getting tail on coin.


Then by total probability theorem,


P (E1) = P (H) P (E1| H) + P (T) P (E1| T)


And P (E2) = P (H) P (E2| H) + P (T) P (E2| T)


Where P (H) = 1/2 ; P (T) = 1/2


P (E1| H) = prob. of getting a sum of 7 on two dice. Here favorable cases are


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


∴ P (E1| H) = 6/36 = 1/6


Also P (E1| T) = prob. if getting ‘7’ numbered card out of 11 cards = 1/11.


P (E2| H) = Prob. of getting a sum of 8 on two dice. Here favorable cases are


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


∴ P (E2| H) = 5/36


P (E2| T) = prob. of getting ‘8’ numbered card out of 11 cards = 1/11


∴ P (E1) = 1/2 x 1/6 + 1/2 x 1/11 = 1/12 + 1/22 = 11 + 6/132 = 17/132


P (E2) = 1/2 x 5/36 + 1/2 x 1/11 = 1/2 [55 + 36/396] = 91/792


Now E1 and E2 are mutually exclusive events therefore


P (E1 or E2) = P (E1) + P (E2) = 17/132 + 91/792


= 102 + 91/792 = 193/792 = 0.2436

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soni singh  (6 years ago)

when we get HEAD the P(getting 7 or 8):- 1/2*(1/6+5/36)=11/72

when we get TAIL the P(getting 7 or 8):- 1/2*(1/11+1/11)=1/11

so P(getting 7 or 8) = 11/72+1/11=0.2436

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