-->

Integrals

Question
CBSEENMA12035688

Two the numbers are selected at random (without replacement) from first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of X. Find the mean and variance of this distribution.

Solution

First six positive integers are {1, 2, 3, 4, 5, 6}
No. of ways of selecting 2 numbers from 6 numbers without replacement = space straight C presuperscript 6 subscript 2 space equals 15 
X denotes the larger of the two numbers, so X can take the values 2, 3, 4, 5, 6.
Probability distribution of X:

X 2 3 4 5 6
P(x) 1/15 2/15 3/15 4/15 5/15

Computation of Mean and Variance:
xi P(X=xi) space straight p subscript straight i straight x subscript straight i
space straight p subscript straight i straight x subscript straight i superscript 2
2 1/15 2/15 4/15
3 2/15 6/15 18/15
4 3/15 12/15 48/15
5 4/15 20/15 100/15
6 5/15 30/15 180/15
    space sum from blank to blank of straight p subscript straight i straight x subscript straight i space equals space 70 over 15 equals 14 over 3 space sum from blank to blank of pix subscript straight i superscript 2 space equals space 350 over 15 equals 70 over 3

space Mean equals sum from blank to blank of straight p subscript straight i straight x subscript straight i space equals space 70 over 15 space equals space 4.67
Variance equals space sum from blank to blank of straight p subscript straight i straight x subscript straight i squared space minus left parenthesis sum from blank to blank of straight p subscript straight i straight x subscript straight i right parenthesis squared space equals 70 over 3 minus 196 over 9 equals fraction numerator 210 minus 196 over denominator 9 end fraction equals 14 over 9

Some More Questions From Integrals Chapter