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Question
Two balls are drawn at random one by one with replacement from an urn containing equal number of red balls and green balls. Find the probability distribution of number of red balls. Also, find the mean of the random variable.
Solution
When we draw 2 balls, we get no red balls,
1 red ball or 2 red balls.
Then, possible values of X are 0, 1, 2.
Let, number of red balls be y and number of green balls be y.
The, total number of balls = 2y
Now, P(X = 0) = P(no red ball)
= (PGG)
= `y/(2y) xx y/(2y)`
= `1/4`
P(X = 1) = P(one red ball)
= P(RG or GR)
= `y/(2y) xx y/(2y) + y/(2y) xx y/(2y)`
= `1/4 + 1/4`
= `1/2`
P(X = 2) = P(two red balls)
= P(RR)
= `y/(2y) xx y/(2y)`
= `1/4`
∴ Probability distribution is
| X | 0 | 1 | 2 |
| P(X) | `1/4` | `1/2` | `1/4` |
Now, mean i.e.,
E(X) = ∑XP(X)
= `0 xx 1/4 + 1 xx 1/2 + 2 xx 1/4`
= `1/2 + 1/2`
= 1.
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