Electronis Discussion
0 votes

A fair coin is tossed till head appears for the first time. The probability that the number of required tosses is odd, is

  1. $\frac{1}{3}$
  2. $\frac{1}{2}$
  3. $\frac{2}{3}$
  4. $\frac{3}{4}$
in Others by (15.7k points)
edited by

1 Answer

0 votes
Probability of getting head = $\frac{1}{2}$

Probability of not getting head = probability of getting tail = $\frac{1}{2}$

Coin is tossed odd number of times untill it’s get first head,

i.e., required outcomes of these trials = {$H,TTH,TTTTH,...$}

So, required probability of getting these outcomes $= \frac{1}{2}+(\frac{1}{2})^3+(\frac{1}{2})^5+….$

$$=\frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{2}{3}$$

Correct Answer: $C$
by (150 points)
Answer:
Welcome to GO Electronics, where you can ask questions and receive answers from other members of the community.
1,044 questions
44 answers
8 comments
42,757 users