Question

A coin is biased so that the head is $3$ times as likely to occur a tail. If the coin is tossed twice, find the probability distribution of number of tails.

• A. $\frac{9}{16},\frac{3}{16},\frac{1}{16}$
• B. $\frac{8}{16},\frac{3}{16},\frac{1}{16}$
  
• C. $\frac{9}{16},\frac{3}{8},\frac{1}{16}$
• D. $\frac{9}{16},\frac{3}{16},\frac{1}{8}$
  

Answer 1

Answer: (C)

$\frac{9}{16},\frac{3}{8},\frac{1}{16}$

Let the probability of getting a tail be p.

Then probability of getting heads be 3p.
$p+3p=1\Rightarrow p=\frac{1}{4}$

Coin is tossed twice. So, n=2.

Probability of 0 tails $=P(X=0)={}_0^{2}C.{\left(\frac{1}{4}\right)}^0{.\left(\frac{3}{4}\right)}^2=\frac{9}{16}$

Probability of 1 tails $=P(X=1)={}_1^{2}C.{\left(\frac{1}{4}\right)}^1{.\left(\frac{3}{4}\right)}^1=\frac{3}{8}$

Probability of 2 tails $=P(X=2)={}_2^{2}C.{\left(\frac{1}{4}\right)}^2{.\left(\frac{3}{4}\right)}^0=\frac{1}{16}$

So, probability distribution is 9/16, 3/8, 1/16.