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Misc 9 - If 4-digit numbers greater than 5,000 are randomly

Misc 9 - Chapter 16 Class 11 Probability - Part 2
Misc 9 - Chapter 16 Class 11 Probability - Part 3 Misc 9 - Chapter 16 Class 11 Probability - Part 4 Misc 9 - Chapter 16 Class 11 Probability - Part 5 Misc 9 - Chapter 16 Class 11 Probability - Part 6 Misc 9 - Chapter 16 Class 11 Probability - Part 7 Misc 9 - Chapter 16 Class 11 Probability - Part 8 Misc 9 - Chapter 16 Class 11 Probability - Part 9 Misc 9 - Chapter 16 Class 11 Probability - Part 10 Misc 9 - Chapter 16 Class 11 Probability - Part 11


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Misc 9 If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (i) the digits are repeated? Digit number greater than 5000 can be formed with either 5 in beginning or 7 in beginning Hence total number of cases = 5 × 5 × 5 × 1 = 125 Hence total number of cases = 5 × 5 × 5 × 1 = 125 Total number of cases = 125 + 125 = 250 But if we get all 0’s after 5, we will get 5000. But we want number greater than 5000 Hence, n(S) = 250 – 1 = 249 Let A be the event that this number is divisible by 5 A number is divisible by 5 in the following cases For (a) Total no of cases = 1 × 5 × 5 × 1 = 25 5 at the end of number place 0 at the end of number place For (b) Total no. of cases = 1 × 5 × 5 × 1 = 25 For (c) Total no of cases = 1 × 5 × 5 × 1 = 25 But if we get all 0’s after 5, we will get 5000. But we want number greater than 5000 Number of cases = 25 – 1 = 24 For(d) Total no. of cases = 1 × 5 × 5 × 1 = 25 Hence, Total numbers divisible by 5 = 25 + 25 + 24 + 25 = 99 So, n(A) = 99 Hence, P(A) = (n(A))/(n(S)) = 99/249 = 𝟑𝟑/𝟖𝟑 Misc 9 If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (ii) the repetition of digits is not allowed? Digit number greater than 5000 can be formed with either 5 in beginning or 7 in beginning Total number of cases = 4 × 3 × 2 × 1 = 24 Total number of cases = 4 × 3 × 2 × 1 = 24 Total number of cases = 24 + 24 = 48 Hence, n(S) = 48 Let B the event that the number greater than 5000 is divisible by 5 Hence it should have either 5 or 0 at unit place A number is divisible by 5 in the following cases For (b) Total number of cases = 1 × 3 × 2 × 1 = 6 Not possible as repetition is not allowed For (c) Total number of cases = 1 × 3 × 2 × 1 = 6 For (d) Total number of cases = 1 × 3 × 2 × 1 = 6 Hence Total number of cases when digit is divisible by 5 = 6 + 6 + 6 = 18 So, n(B) = 18 Hence, P(B) = (n(B))/(n(S)) = 18/48 = 𝟑/𝟖

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.