Suppose you have 3 bags. Two of them contain a single $10 bill, and the third contains a single $5 bill. Suppose you pick one of these bags uniformly at random. You then add a $5 bill to the bag, so it now contains two bills. The bag is shaken, and you randomly draw a bill from the bag without looking into the bag. Suppose it turns out to be a $5 bill. If a you draw the remaining bill from the bag, what is the probability that it, too, is a $5 bill

1/2

Step-by-step explanation:

Number of bags = 3

number of bags with $10 bill initially = 2

number of bags with $5 bill initially = 1

assume :

event you pick a $5 bill at first draw = A

event you pick a $5 bill at second draw = B

hence : P ( A n B ) = 1/3 * 1 = 1/3

P( A ) = ( 1/3 * 1 ) + ( 1/3 * 1/2 + 1/3 * 1/2 )  = 2/3

therefore P( that the second drawn bill is $5 )

P( B | A ) = P(A n B ) / P ( A )

              = (1/3) / (2/3) = 1/2

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.

step-by-step explanation:

se a cavasery6