53. Find the probability distribution of the number of success in two tosses of a die where a success is defined as:

(i) Number greater than 4.

(ii) Six appears on at least one die.

<p><strong>53. </strong>When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.</p><p>Let X be the random variable, which represents the number of successes.</p><p><strong> (i)</strong> Here, success refers to the number greater than 4.</p><p>P (X = 0) = P (number less than or equal to 4 on both the tosses) = 4/6 X 4/6 = 4/9</p><p>P (X = 1) = P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P (number greater than 4 on first toss and less than or equal to 4 on second toss)</p><p>= 4/6 X 2/6 + 4/6 X 2/6 = 4/9</p><p>P (X = 2) = P (number greater than 4 on both the tosses)</p><p>&nbsp;= 2/6 X 2/6 = 1/9</p><p>Thus, the probability distribution is as follows.</p><table><tbody><tr><td><p>X</p></td><td><p>0</p></td><td><p>1</p></td><td><p>2</p></td></tr><tr><td><p>P (X)</p></td><td><p>4/9</p></td><td><p>4/9</p></td><td><p>1/9</p></td></tr></tbody></table><p><strong> (ii)</strong>&nbsp;Here, success means six appears on at least one die.P (Y = 0 ) = P (six appears on&nbsp; none of the dice) = 5/6 x&nbsp; 5/6 = 25/36</p><p>P (Y = 1) = P (six appears on at least one of the dice) = 1/6 x 5/6 + 5/6 x 1/6 + 1/6 x 1/6 = 11/36</p><p>Thus, the required probability distribution is as follows.</p><table><tbody><tr><td><p>Y</p></td><td><p>0</p></td><td><p>1</p></td></tr><tr><td><p>P (Y)</p></td><td><p>25/36</p></td><td><p>11/26</p></td></tr></tbody></table>

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