Let x denote the number of doublets in three throws of a pair of dice with the following probability distribution.
|
X |
0 |
1 |
2 |
3 |
|
P(x) |
$\frac{25}{72}k$ |
$\frac{15}{72}k$ |
$\frac{3}{72}k$ |
$\frac{1}{360}k$ |
If value of k is equal to $\frac{m}{n}, gcd(m, n) = 1$, then $m+n$ is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8
Given: Probability distribution of number of doublets in 3 throws of a pair of dice:
| $X$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(X)$ | $\frac{25}{72}k$ | $\frac{15}{72}k$ | $\frac{3}{72}k$ | $\frac{1}{360}k$ |
Use total probability = 1:
$\frac{25}{72}k + \frac{15}{72}k + \frac{3}{72}k + \frac{1}{360}k = 1$
$k \left( \frac{25 + 15 + 3}{72} + \frac{1}{360} \right) = 1$
$k \left( \frac{43}{72} + \frac{1}{360} \right) = 1$
Take LCM of 72 and 360 = 360
$\frac{43}{72} = \frac{215}{360}$, so:
$k \left( \frac{215 + 1}{360} \right) = 1 \Rightarrow k \cdot \frac{216}{360} = 1$
$k = \frac{360}{216} = \frac{5}{3}$
So $k = \frac{m}{n} = \frac{5}{3} \Rightarrow m + n = {8}$