An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on two faces is noted. If the result is a tail, a card from a well-shuffled pack of 11 cards numbered 2, 3, 4, ..., 12 is picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8?

Answer

A, B, C

Explanation

Let, `E_(1)`= the event noted number is 7
`E_(2)`= the event noted number is 8
H = getting head on coin
T = be getting tail on coin
`:." By law of total probability,"`
`P(E_(1))=P(H).P(E_(1)//H)+P(T).P(E_(1)//T)`
and `P(E_(2))=P(H).P(E_(2)//H)+P(T).P(E_(2)//T)`
where, `P(H)=1//2=P(T)`
`P(E_(1)//H)`= probability of getting a sum of 7 on two dice Here, favourable cases are
`{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)}.`
`:." "P(E_(1)//H)=(6)/(36)=(1)/(6)`
Also, `P(E_(1)//T)`= probability of getting 7 numbered card ont of 11 cards
`=(1)/(11)`
`P(E_(2)//H)`= probabilitiy of getting a sum of 8 on two dice Here, favourable cases are
`{(2,6),(6,2),(4,4),(5,3),(3,5)}.`
`:." "P(E_(2)//H)=(5)/(36)`
`P(E_(2)//T)`= probability of getting '8' numbered card out of 11 cards
`=1//11`
`:." "P(E_(1))=((1)/(2)xx(1)/(6))+((1)/(2)xx(1)/(11))=(1)/(12)+(1)/(22)=(17)/(132)`
and `" "P(E_(2))=((1)/(2)xx(5)/(36))+((1)/(2)xx(1)/(11))`
`=(1)/(2)((91)/(396))=(91)/(729)`
Now, `E_(1)` and `E_(2)` are mutually exclusive events.
Therefore,
`P(E_(1)" or "E_(2))=P(E_(1))+P(E_(2))=(17)/(132)+(91)/(792)=(193)/(792)`