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?

Explanation

`{:(E_(1)-="Number noted is 7",),(E_(2)-="Number noted is 8",),(H-="Getting tail on coin",),(T-="Getting tail on coin",):}`
Then, by total probability theorem,
`P(E_(1))=P(H)P(E_(1)//H)+P(T)P(E_(1)//T)`
`P(E_(2))=P(H)P(E_(2)//H)+P(T)P(E_(2)//T),`
where P (H) =(1/2), P(T) = (1/2), and `P(E_(1)//H)` is the probability 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))} Therefore,
`P(E_(1)//H)=(6)/(36)=1/6`
Also, `P(E_(1)//T)` is the probability of getting '7' numbered card out of 11 cards. Therefore,
`P(E_(1)//T)=1//11`
`P(E_(2)//H)` is the probability of getting a sum of 8 on two dice. Here, favorable cases are `{(2,6)(6,2)(4,4),(5,3),(3,5)}` Therefore
`P(E_(2)//H)=5/36`
The probability of getting '8' numbered card out of 11 cards is `P(E_(2)//T)=1//11.` Therefore,
`P(E_(1))=1/2xx1/6+1/2xx1/11=1/12=(11+6)/(132)=17/132`
`P(E_(2))=1/2xx5/36+1/2xx1/11=1/2[(55+36)/(396)]=91/792`
Now, `E_(1)and E_(2)` are matually exclusive events, therefore,
`P(E_(1)uuE_(2))=P(E_(1))+P(E_(2))`
`=17/132+91/792=(102+91)/(792)=(193)/(792)`