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

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=12Sum 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