A biased coin in which the head is 3 times as likely to occur as tail, is tossed twice. If the probability distribution of the number of tails is given by:
|
X |
0 |
1 |
2 |
|
P(X = x) |
$\frac{9}{16}$ |
m |
n |
The value of $\frac{m}{n}$ is:
Answer & explanation
Correct answer: option 2
Let x be the random variable as ‘Number of tails’.
∴ P.D table
|
X |
0 |
1 |
2 |
|
P(X=x) |
$\frac{9}{16}$ |
$\frac{3}{8}$ |
$\frac{1}{16}$ |
∴ x = 0, 1, 2
x = 0 means no tail
x = 1 means 1 tail
x = 2 means both tail
$P(H)=\frac{3}{4}$, $P(T)=\frac{1}{4}$
S = {HH, HT, TH, TT}
P(x = 0) = P(HH) = $\frac{3}{4}.\frac{3}{4}=\frac{9}{16}$
P(x = 1) = P(HT or TH) = P(HT) + P(TH) = $\frac{3}{4}.\frac{1}{4}+\frac{1}{4}.\frac{3}{4}=\frac{6}{16}=\frac{3}{8}$
P(x = 2) = P(TT) = $\frac{1}{4}.\frac{1}{4}=\frac{1}{16}$
∵ P(x = 1) = m = $\frac{3}{8}$
& P(x = 2) = n = $\frac{1}{16}$
$\frac{m}{n}=\frac{\frac{3}{8}}{\frac{1}{16}}$
$=\frac{3}{8}×\frac{16}{1}=6$