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Question
Two biased dice are thrown together. For the first die P(6) = `1/2`, the other scores being equally likely while for the second die, P(1) = `2/5` and the other scores are equally likely. Find the probability distribution of ‘the number of ones seen’.
Solution
Given that: for the first die, P(6) = `1/2`
And `"P"(bar6) = 1 - 1/2 = 1/2`
⇒ P(1) + P(2) + P(3) + P(4) + P(5) = `1/2`
But P(1) = P(2) = P(3) = P(4) = P(5)
∴ 5.P(1) = `1/2`
⇒ P(1) = `1/10` and `"P"(bar1) = 1 - 1/10 = 9/10`
For the second die, P(1) = `2/5` and `"P"(bar1) = 1 - 2/5 = 3/5`
Let X be the number of one’s seen
∴ X = 0, 1, 2
⇒ P(X = 0) = `"P"(bar1)."P"(bar1)`
= `9/10 * 3/5`
= `27/50`
= 0.54
P(X = 1) = `"P"(bar1)*"P"(1) + "P"(1)*"P"(bar1)`
= `9/10*2/5 + 1/10*3/5`
= `(18 + 3)/50`
= `21/50`
= 0.42
P(X = 2) = `"P"(1)_"I" * "P"(1)_"II"`
= `1/10*2/5`
= `2/50`
= 0.04
Hence, the required probability distribution is
| X | 0 | 1 | 2 |
| P(X) | 0.54 | 0.42 | 0.04 |
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