We are given that the probability of choosing Wn is: ‌P(Wn)=2P(Wn−1)‌ for ‌n>1 ‌⇒P(W1)=p,P(W2)=2p,P(W3)=4p,...,P(Wn)=2n−1p
To find the value of p, we use:
120
∑
n−1
P(Wn)=1‌ (total probability) ‌ So, p(1+2+22+...+2119)=1 p(2120−1)=1⇒p=‌
1
2120−1
Thus, P(Wn)=‌
2n−1
2120−1
....(i)
Since the first letter of CDBEA is C Words starting with A:4!=24 Words starting with B:4!=24 ‌CA−−−:3!=6 ‌CB_−_:3!=6 ‌CDA−−:2!=2 ‌CDBA−:1!=1 Total before CDBEA=63 Position of CDBEA =64‌th ‌ Putting in (i) P(CDBEA)=P(W64)=‌