Milicevic3306
asked
in Probability and Statistics
Mar 25, 2018
retagged
Mar 3, 2021
by Lakshman Patel RJIT

79 views
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: $C$

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: $C$