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mathmath333
 one year ago
Probability question
mathmath333
 one year ago
Probability question

This Question is Closed

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0If 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?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3OK, 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.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0yes i m confused in that

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3dw:1439703404482:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0will 2*4*3*2 will also include the number 5000

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3The 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Because we are actually decreasing the number of choices in each blank as we're not allowed to repeat digits, hai na?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, so this is the total number of permutations. For divisibility by 5, the last digit should either be 0 or 5.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Great. Explain how you found that.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439704058607:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3My 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.

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0isn't there a formula for this? and just multiply them.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3There usually is, but here we have lots of constraints. :\

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0why did u add both cases (thousand digit 7) and ( thousand digit 0, 5)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3The 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.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439704691337:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3No, 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.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but repetition is not allowed

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Yes, and we're not repeating the digits in a number like 5730.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3You 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3dw:1439705072309:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Maine 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I hope maine confuse na kar diya ho. Agar kaheen samajh na aa raha ho, toh please ask. :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3But that is only for the case where the last (units') digit is zero.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439705470580:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0for second case 1*3*2*2 ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3What's the second case? Units' digit 5?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3OK, 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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3OK, so look through both cases and try to spot the difference between the two.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.05 is not in thousand's place

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Yes, 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.
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