An unbiased coin is tossed. If the outcome is a head, then a pair of unbiased dice is rolled and the sum of the number obtained on them is noted. If the toss of the coin results in tail, then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, ...., 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is

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)`