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mathmath333

  • one year ago

Probability question

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  1. mathmath333
    • one year ago
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    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 the repetition of digits is not allowed?

  2. ParthKohli
    • one year ago
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    OK, so don't think of it as a probability question. You just have to count the total number of permutations and the number of permutations where the number is divisible by 5. Then you divide them.

  3. mathmath333
    • one year ago
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    yes i m confused in that

  4. ParthKohli
    • one year ago
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    |dw:1439703404482:dw|

  5. mathmath333
    • one year ago
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    will 2*4*3*2 will also include the number 5000

  6. ParthKohli
    • one year ago
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    The first digit is 5 or 7 because we want the number to be greater than 5,000. There are two choices. Now suppose we choose one number. For the second blank, there are four choices out of the five numbers 0, 1, 3, 5, 7 as we already chose one for the first blank. For the subsequent blanks, we keep decreasing by one choice. And no, this wouldn't include 5000 or any other number where repetition of digits is happening.

  7. ParthKohli
    • one year ago
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    Because we are actually decreasing the number of choices in each blank as we're not allowed to repeat digits, hai na?

  8. mathmath333
    • one year ago
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    ya correct

  9. ParthKohli
    • one year ago
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    Yeah, so this is the total number of permutations. For divisibility by 5, the last digit should either be 0 or 5.

  10. mathmath333
    • one year ago
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    =2*2*1*2

  11. mathmath333
    • one year ago
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    ^for numerator

  12. ParthKohli
    • one year ago
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    Great. Explain how you found that.

  13. mathmath333
    • one year ago
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    |dw:1439704058607:dw|

  14. ParthKohli
    • one year ago
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    My finding is this way: Consider the two cases for divisibility by 5. _ _ _ 5 _ _ _ 0 In the first case, the first digit has to be 7 in order for you to get a number greater than 5000. So we have to fix 7 as the first digit. For the middle two digits, we have 3 * 2 = 6 ways. So there are 6 ways for the first case. In the second case, the first digit can be 5 or 7 (two choices), then the second digit has 3 choices (since two out of five digits have been counted), and the fourth has 2 choices. So in this case, we can have 2 * 3 * 2 = 12 ways. Therefore, the total number of ways should be 12 + 6 = 18 for the numerator.

  15. abb0t
    • one year ago
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    isn't there a formula for this? and just multiply them.

  16. ParthKohli
    • one year ago
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    There usually is, but here we have lots of constraints. :\

  17. mathmath333
    • one year ago
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    why did u add both cases (thousand digit 7) and ( thousand digit 0, 5)

  18. ParthKohli
    • one year ago
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    The first case is for units' digit 5. The second is for units' digit 0. If we find the number of permutations in each case satisfying the given condition, then we can simply add them to find the total number of permutations satisfying the given condition.

  19. mathmath333
    • one year ago
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    |dw:1439704691337:dw|

  20. ParthKohli
    • one year ago
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    No, not totally. Aisa nahi hai ki the five and seven couldn't go to the second and third digits. Like the number 5730 also satisfies the condition but you're not counting it.

  21. mathmath333
    • one year ago
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    but repetition is not allowed

  22. ParthKohli
    • one year ago
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    Yes, and we're not repeating the digits in a number like 5730.

  23. ParthKohli
    • one year ago
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    I encourage you to not write the specific numbers allowed because they're dependent on the previous entries in the previous blanks. You've written 1, 3 for the third blank but it actually depends on how you've filled the previous blanks. If you've already filled 1, then you can't write 1. Also if you didn't fill 7 in the first and second blanks, then you can fill it in the third.

  24. ParthKohli
    • one year ago
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    You should look at it as the number of possible choices for each blank. It's simpler to look at it that way because it just decreases by 1 when repetition is not allowed.

  25. ParthKohli
    • one year ago
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    |dw:1439705072309:dw|

  26. ParthKohli
    • one year ago
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    Maine pehle blank ke liye 5/7 isliye likha kyunki: 1. Wahaan pe 5 ya 7 hi allowed hai. 2. Main wahaan se filling shuru kar raha hoon, toh it is independent of the second, third, and fourth blanks.

  27. ParthKohli
    • one year ago
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    I hope maine confuse na kar diya ho. Agar kaheen samajh na aa raha ho, toh please ask. :)

  28. mathmath333
    • one year ago
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    2*3*2*1 ?

  29. ParthKohli
    • one year ago
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    Yes.

  30. ParthKohli
    • one year ago
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    But that is only for the case where the last (units') digit is zero.

  31. mathmath333
    • one year ago
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    |dw:1439705470580:dw|

  32. mathmath333
    • one year ago
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    for second case 1*3*2*2 ?

  33. ParthKohli
    • one year ago
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    What's the second case? Units' digit 5?

  34. mathmath333
    • one year ago
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    yes 5 or 0

  35. ParthKohli
    • one year ago
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    OK, so the favourable cases are the ones where the number is divisible by 5. By the divisibility test, there are only two ways this can happen: either the units' digit is 5, or 0. Now do you know why we can't count them together?|dw:1439705814387:dw|

  36. mathmath333
    • one year ago
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    no repeat

  37. ParthKohli
    • one year ago
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    OK, so look through both cases and try to spot the difference between the two.

  38. mathmath333
    • one year ago
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    5 is not in thousand's place

  39. ParthKohli
    • one year ago
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    Yes, that's in the case where 5 is the unit place. Because we cannot place 5 in the thousands' place so the only choice we now have is 7.

  40. mathmath333
    • one year ago
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    ok

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