Math, asked by keeku1357, 1 year ago

Two numbers are selcted at random form the fist six positive intefgers let x denote the larger of the two numbers ontained


Anonymous: Are the two numbers selected with replacement so there are 36 equally likely outcomes? Or are the two numbers selected without replacement so there are 15 equally likely outcomes (in particular, doubles do not occur)? And then what about X? Do you need the expected value of X? The variance?

Answers

Answered by sashanksaimanikanta
0

The first six positive integers are 1, 2, 3, 4, 5, 6.

We can select the two positive numbers in 6 × 5 = 30 different ways.

Out of this, 2 numbers are selected at random and let X denote the larger of the two numbers.

Since X is the large of the two numbers, X can assume the value of 2, 3, 4, 5 or 6.

P (X =2) = P (larger number is 2) = {(1,2) and (2,1)} = 230

P (X = 3) = P (larger number is 3) = {(1,3), (3,1), (2,3), (3,2)} = 430

P (X = 4) = P (larger number is 4) = {(1,4), (4,1), (2,4), (4,2), (3,4), (4,3)} = 630

P (X = 5) = P (larger number is 5) = {(1,5), (51,), (2,5), (5,2), (3,5), (5,3), (4,5), (5.4)} = 830

P (X = 6) = P (larger number is 6) = {(1,6), (6,1), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4), (5,6), (6,5)} = 1030

Given the above probability distribution, the expected value or the mean can be calculated as follows:

Mean = ∑(Xi×P(Xi)) = 2×230+3×430+4×630+5×830+6×1030=4+12+24+40+6030=14030=143

Similar questions