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

• 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
  
• 8 years agoHelpfull: Yes(0) No(1)
• 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
• 7 years agoHelpfull: Yes(0) No(1)