1. A bag contains five white balls and four black balls. Your goal is to draw two black balls.
a)You draw two balls at random. What is the probability that they are both black?
b)You draw two balls at random. Once you have drawn two balls, you put back any white balls, and redraw so that you again have two drawn balls. What is the probability that you now have two black balls? (Include the probability that you chose two black balls on the first draw.)
2. From a standard deck of 52 cards, the four Aces, four 's, and four 's are drawn, forming a smaller deck of cards. All cards are dealt at random to four players, so that each player gets three cards. What is the probability that each player's hand consists of an Ace, a , and a ?
1. (a) FIRST BALL: So the probability of first drawing a black ball or a white ball is both 5/10.
SECOND BALL: The probability of getting a second ball is 4 out of 9 (because you already took a ball), so for both it's 4/9.
To find the probability, of both, you can multiply 5/10 * 4/9, = 2/9. Which is the answer.
(b) You must consider the probability that drawing BB, BWB, WWB, or WWBB. This gives you a probability of 2/9 + 5/81 + 12/81 + 6/81 = 41/81.
Hi, I think you read the question wrong. For part a), there are 9 balls in total, not 10. Therefore, both of your answers were wrong. However, thank you so much for trying to help! (If you would like to reattempt the questions or if anyone else could, that would be greatly appreciated!)
Also, for 2., the question should be:
From a standard deck of 52 cards, the four Aces, four 2's, and four 3's are drawn, forming a smaller deck of 12 cards. All 12 cards are dealt at random to four players, so that each player gets three cards. What is the probability that each player's hand consists of an Ace, a 2, and a 3?
Sorry about the formatting errors. I will double check them in future.