Let p= probability of getting a tail in a single trial =1/2
n= number of trails =100
and X= number of tails in 100 trials.
We have
P(X=r)=100Crprqn−r
=100Cr(12)r(12)100−r=100Cr(12)100
Now,
P(X=1)+P(X=3)+....+P(X=49)
=100C1(12)100+100C3(12)100+....+100C49(12)100
=(100C1+100C3+...+100C49)(12)100
But 100C1+100C3+...+100C99=299
Also, 100C99=100C1,
100C97=100C3,...100C51=100C49
Thus,
2(100C1+100C3+...+100C49)=299
⇒100C1+100C3+...+100C49=298
∴ Probability of required event =2982100=14.