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Question
A fair die is tossed twice. If the number appearing on the top is less than 3, it is a success. Find the probability distribution of number of successes.
Solution
Let X denote the event of getting a number less than 3 (1 or 2) on throwing the die. Then, X can take the values 0, 1 and 2.
Now,
\[P\left( X = 0 \right) = \frac{16}{36} = \frac{4}{9}\]
\[P\left( X = 1 \right) = \frac{16}{36} = \frac{4}{9} \]
\[P\left( X = 2 \right) = \frac{4}{36} = \frac{1}{9}\]
Thus, the probability distribution of X is given by
| X | P(X) |
| 0 |
\[\frac{4}{9}\]
|
| 1 |
\[\frac{4}{9}\]
|
| 2 |
\[\frac{1}{9}\]
|
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