Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of X. Find the mean and the variance of this distribution.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Mean: $\frac{14}{3}$, Variance: $\frac{14}{9}$
The number of ways of choosing two integers (without replacement) from the first six positive integers = ${^6C}_2 = 15$, so the sample space S has 15 equally likely outcomes. These outcomes are:
1, 2; 1, 3; 1, 4; 1, 5; 1, 6; 2, 3; 2, 4; 2,5;
2, 6; 3, 4; 3, 5; 3, 6; 4, 5; 4, 6; 5, 6.
As the random variable X denotes the larger of the two numbers, X can take values 2, 3, 4, 5, 6.
(Because 1 is not larger than any number from 1 to 6)
Note that in the sample space S, we have
|
Larger of two numbers |
Number of outcomes |
|
2 |
1 |
|
3 |
2 |
|
4 |
3 |
|
5 |
4 |
|
6 |
5 |
$P(X = 2) =\frac{1}{15}= P(X = 3) = \frac{2}{15}, P(X = 4) =\frac{3}{15}$
$P(X = 5)=\frac{4}{15},P(X = 6) =\frac{5}{15}$.
The probability distribution of X is:
|
$X$ |
2 |
3 |
4 |
5 |
6 |
|
$P(X)$ |
$\frac{1}{15}$ |
$\frac{2}{15}$ |
$\frac{3}{15}$ |
$\frac{4}{15}$ |
$\frac{5}{15}$ |
Mean = $μ = Σp_ix_i = \frac{1}{15}= (1 × 2+2×3+3×4 + 4×5 +5 × 6)$
$=\frac{1}{15}(2+6+12+20+30)=\frac{70}{15}=\frac{14}{3}$.
Now $Σp_i{x_i}^2=\frac{1}{15}=(1 × 2^2+2×3^2+3×4^2 + 4×5^2 +5 × 6^2)$
$=\frac{1}{15}(4+18+48+100+180)=\frac{350}{15}=\frac{70}{3}$
∴ Variance = $Σp_i{x_i}^2-μ^2=\frac{70}{3}-(\frac{14}{3})^2=\frac{70}{3}-\frac{196}{9}=\frac{14}{9}$.