There are four balls in a hat: a blue one, a white one, and two red ones. Now I draw simultaneously two balls, look at them, and announce that at least one of them is red.
What is the chance that the other is red as well?
In the card equation, the proposed solution is correct. 25/51. Per the instruction (if the announcement is truthful), if two black cards are drawn, the draw would not count, the two cards replaced, reshuffled, and two more cards drawn.
Likewise with the marbles, if the blue and white marbles are drawn, the result would not be included with this exercise.
1blue
2white
3red
4red
There are six combinations.
12
13
14
23
24
34
The first combo is discarded since it does not contain a red marble. Of the five remaining, only one includes two red marbles (34).
Probability, 20%.
P(AB) = P(BA) * P(A) / P(B)
P(BA) = 1
P(A) = 1/6
P(B) = 5/6

P(AB) = 1/5

Posted by hoodat
on 20170520 20:59:15 