Question

Bob has a coin cup with four $1 tokens and two $5 tokens in it. He also has two $10 tokens and one $25 token in his pocket. He randomly draws a token from the cup, and then randomly draws a token from his pocket. What is the probability that he will draw $11 in tokens?

Answer 1

do you have answer choices?

Answer 2

Yes, 1/9, 2/9, 2/3, and 4/9

Answer 3

well to get $11 from 2 coins, he can only draw a $1 and a $10

Answer 4

and there are 9 coins in total

Answer 5

so \(\dfrac{1}{9}\)

Answer 6

I appreciate your help, but it was 4/9

Answer 7

THANK YOU! :)

Answer 8

would you like to see why you got the problem wrong?

Answer 9

YES, PLEASE!

Answer 10

as as said, the only combination that can yield $11 is a $1 and a $10. The two drawing are separate though, and there are more than one of each kind of coin in each place (cup and pocket)
what are the odds that he draws a $1 from the cup?

Answer 11

was said*

Answer 12

1/4?

Answer 13

not quite
odds of drawing a $1 token from the cup is the number of $1 tokens divided by the total number of tokens in the cup
how many $1's are in the cup
how many tokens total?

Answer 14

I got 41 tokens with 4 $1 coins, so 4/41?

Answer 15

No, 4/9

Answer 16

no, just in the cup
there are 4 $1 tokens, and how many tokens total *in the cup*?

Answer 17

2/3 and/ or 4/6

Answer 18

good
now for the pocket, what are the odds of drawing a $10 ?

Answer 19

2/3

Answer 20

good
now, the drawing from the cup and pocket don't affect each other, which means they are *independent*
if we have two independent events, and we want to find the odds of both of them occuring, we multiply the probabilities

i.e. probability of drawing 11 = probability of drawing a 1 times the probability of drawing a 10

What is the product of those probabilities?