A coin whose faces are marked 3 and 5 is tossed 4 times. The probability that the sum of the numbers thrown is greater than 15 is
1116
516
58
116
let ' r ' denote the number of times 3 appears n=4,r=0,1,2,3,4
'P′ be the probability of a coin showing 3
⇒p=12,p+q=1⇒q=12
Sum is greater than 15
At least 2 coins show 5At most 2 coins show 3P0≤X=r≤2
=PX=0+PX=1+PX=2 (∵PX=r=nCrprqn−r)
=4C0.P0.q4+4C1 .p1q3+4C2.p2q2
=1241+4+6
=1116