AMC 8 Daily Practice Round 1

Complete problem set with solutions and individual problem pages

Problem 29 Medium

Vendors often run lottery games at the school gate. One vendor has a black bag containing 50 balls of different colors: 1 red, 2 yellow, 10 green, and the rest white. After mixing them thoroughly, the rule is: each draw costs 2 dollars for 1 ball.

Drawing a red ball wins a prize worth 8 dollars.

Drawing a yellow ball wins a prize worth 5 dollars.

Drawing a green ball wins a prize worth 2 dollars.

Drawing a white ball wins no prize.

If you spend 4 dollars to draw 2 balls at the same time, what is the probability of obtaining a prize worth 10 dollars?

  • A.

    \frac{18}{1225}

  • B.

    \frac{11}{1225}

  • C.

    \frac{121}{225}

  • D.

    \frac{11}{245}

  • E.

    \frac{2}{245}

Answer:B

There are two cases for winning a prize worth 10 dollars:

Case 1: Drawing one 8-dollar prize and one 2-dollar prize, i.e., one red ball and one green ball. The probability of drawing a red ball first and then a green ball, or a green ball first and then a red ball, is the same, giving \frac{1 \times 10}{50 \times 49} \times 2 = \frac{2}{245}.

Case 2: Drawing two yellow balls. The probability of drawing the first yellow ball is \frac{2}{50}, and the probability of drawing the second yellow ball is \frac{1}{49}. Thus, \frac{2}{50} \times \frac{1}{49} = \frac{1}{1225}.

Therefore, the total probability is \frac{2}{245} + \frac{1}{1225} = \frac{11}{1225}.