53. If 4-digit numbers greater than 5,000 are randomly formed from the digits0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (i) The digits are repeated? (ii) The repetition of digits is not allowed?

<p><strong>53.</strong>&nbsp; (a)<strong> </strong>No. of ways of forming a four-digit number greater than 5000 from the given digit 0, 1, 3, 5, 7. and digit repetition is allowed can be done in such a way either 5 or 7 and occupy the thousands' place and any of the digits 0, 1, 3, 5, 7 can occupy the remaining 3 places.</p><p>Hence, the required no. of ways = (2* 5 * 5 * 5) - 1</p><p>= 250 - 1 = 249</p><p>Here 1 is subtracted because 5000 which can be formed by the permutation of the given digits is not allowed</p><p>Hence, n (s) = 249.</p><p>Similarly, in order to formed a number divisibleby 5 we need to have either 0 or 5 in the one place.</p><p>The required number of ways = 2* 5 *5 *2 - 1</p><p>= 100 - 1</p><p>= 99.</p><p>Hence, no. of favourable went, n (E) = 99.</p><p>Probability that four digit no. greater than 5000 formed by digits 0, 1, 3, 5, 7 (when repetition is) allowed and is divisible by 5 is</p><p>P (E) =&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mi>n</mi><mo stretchy="false"> (</mo><mi>E</mi><mo stretchy="false">)</mo></mrow><mrow><mi>n</mi><mo stretchy="false"> (</mo><mi>s</mi><mo stretchy="false">)</mo></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>9</mn><mn>9</mn></mrow><mrow><mn>2</mn><mn>4</mn><mn>9</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>3</mn><mn>3</mn></mrow><mrow><mn>8</mn><mn>3</mn></mrow></mfrac></mrow></math></span></p><p> (B) When repetition is not allowed</p><p>We can have one of either 7 or 5 on the thousands' place. Then the remaining four other digits can occupy the hundreds, place and then the remaining 3 other digits can occupy tens' places and lastly the remaining 2 other digits can occupy the ones' place.</p><p>So, the required number of ways = 2* 4 *3* 2</p><p>n (s) = 48</p><p>Now, number of 4-digit numbers starting with 5 and divisible by 5 will have 5 on thousands' place and 0 on ones' place then the remaining 3 other digits can fill the hundreds' place and then the remaining 2 other digits can fill the tens' place. So, the required number of ways = 1 *3 *2* 1</p><p>= 6</p><p>Similarly, 4-digit number starting with 7 divisible by 5 will have 7 on the thousand's place and (one of) 0 or 5 on ones' place. Then the remaining.</p><p>3 other digits can fill the hundreds' place and then the remaining 2 other digits can fill the tens' place.</p><p>Hence the required no. of ways =1*3* 2 *2=12</p><p>So, Total no. of four digit number divisible by 5=6+12=18</p><p>Probability that four digit no. greater than 5000 formed</p><p>by digits 0,1, 3, 5, 7when repetition is not allowed and is divisible by 5 is</p><p><span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>1</mn><mn>8</mn></mrow><mrow><mn>4</mn><mn>8</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>3</mn></mrow><mrow><mn>8</mn></mrow></mfrac><mo>.</mo></mrow></math></span></p>

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