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# help pls

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Krishanu and Shaunak each pick an integer at random between 1 and 10, inclusive. What is the probability that the product of their numbers is more than 10? Express your answer as a common fraction.

Dec 1, 2019

#1
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10 C 2 = 45. Or:

(2, 3, 4, 5, 6, 6, 7, 8, 8, 9, 10, 10, 12, 12, 14, 15, 16, 18, 18, 20, 20, 21, 24, 24, 27, 28, 30, 30, 32, 35, 36, 40, 40, 42, 45, 48, 50, 54, 56, 60, 63, 70, 72, 80, 90) = 45

The products over 10 = 33. Therefore: The probability is 33 / 45 =11 / 15

Dec 1, 2019
#2
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Hi Guest,

You have assumed that the two must choose different digits.  I do not think that is the case.

To answer I drew a 10 by 10 grid. Like a times table grid.

There are 100 pairs of numbers in the grid. That is the sample space.

Then I ticked of the pairs that met the condition. I found 73 of them

So the answer I go was   73/100

Dec 1, 2019
#3
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To make this easier, we can find the probability of the numbers being less than or equal to 10, and then subtract that from 1.

Combinations:

We can find that all 1's work.

1 x 1, 1 x 2, 1 x 3, 1 x 4, ... , 1 x 10

Same with the other way

1 x 1, 2 x 1, 3 x 1, 4 x 1, ... , 10 x 1

Now we subtract 1 because we counted 1 x 1 twice, to get a total of 19.

Now for 2's:

2 x 2, 2 x 3, 2 x 4, 2 x 5

2 x 2, 3 x 2, 4 x 2, 5 x 2

Subtract 1 and get 7.

Now 3's:

3 x 3

3 x 3

Subtract 1 and get 1.

4 or higher won't work because it's greater than 10, and we already counted values like 4 x 1 and 4 x 2.

19+7+1 = 27

There are a total of 100 possibilities

So 1-27/100 = 73/100.

You are very welcome!

:P

Dec 1, 2019