# Probability and divisible by questions help please!!?

A.) Five counterfeit coins are mixed with nine authentic coins. If two coins are drawn at random, what is the probability that one coin is authentic and one is counterfeit?

B.) What is the sum of the integers between 1 and 300 that are divisible by 11, 13 or both?

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A) First, you need the total number of coins. Add 5+9=14

so you have 14 coins

You have two possibilities for drawing 1 real and 1 counterfeit coin

you can draw the counterfeit coin first, a 5/14 chance, then draw a real one, a 9/13 chance, or reverse that and draw a real one first, 9/14, then a fake, 5/13. Now multiply the fractions together, because you draw twice.

5/14 x 9/13=.247 or 24.7%

9/14 x 5/13=.247 or 24.7%

Now you add these.

24.7+24.7=49.4% of the time this will happen

B) An integer is a whole number.

integers divisible by 11: 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143, 154, 165, 176, 187, 198, 209, 220, 231, 242, 253, 264, 275, 286, 297

divisible by 13: 13, 26, 39, 52, 65, 78, 91, 104, 117, 130, 143, 156, 169, 182, 195, 208, 221, 234, 247, 260, 273, 286, 299

If any numbers are repeated, only use that number once, not twice

in this case, 143 and 286 are both divisible by 11 and 13, but you only add them once.

Add all number:

11+22+33+44+55+66+77+88+99+110+121+132+143+154+165+176+187+198+209+220+231+242+253+264+275+286+297+13+26+39+52+65+78+91+104+117+130+(already have 143 so jump to)156+169+182+195+208+221+234+247+260+273+(again, already have 286, so)299

All of these end up equaling 7317

A) (5×9)/(14×13) + (9×5)/(14×13)

B)

(a) sum integers 1:300 is 300×301/2

(b) [300/11]=27. (where [] = "integer part of")

sum integers 1:300 div by 11 is thus 11*(27*28)/2

(c) [300/13] = 23

sum integers 1:300 div by 13 is thus 13*(23*24)/2

(d) sum integers divisivle by 11 AND 13 is 143*(2*3)/2

required answer is a - b - c + d (add d because we subtracted it twice)