A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is mn, (m, n)=1, then m+n is equal to :
- 53
- 55
- 107
- 105
Answer
(3) 107
Solution
He gets exactly 30 points if his running total hits 30 at some stage, i.e. the first few throws give 10h+5t=30 (h heads, t tails):
2h+t=6 (h, t)=(0, 6), (1, 4), (2, 2), (3, 0)
A particular order of h+t throws has probability ( 12 )h+t, and the heads can be placed in h+tCh ways:
(0, 6): 1 164= 164
(1, 4): 5C1 132= 1064
(2, 2): 4C2 116= 2464
(3, 0): 1 18= 864
These cases are mutually exclusive.
P= 1+10+24+864= 4364
m=43, n=64 (coprime), m+n=107
Answer: (3)