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Mathematics, four units · Ages 16–17

Probability of Draws Without Replacement

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For 11th-grade math students. Calculate the probability for each scenario where items are drawn without replacement. Give answers as simplified fractions.

  1. A bag contains 4 red balls and 6 green balls. You draw one ball, keep it, and draw a second. What is the probability that both balls are red?

  2. A drawer holds 8 black socks and 4 blue socks. You pull out two socks at random without replacement. What is the probability that both are black?

  3. A box contains 5 green tokens, 3 yellow tokens, and 2 white tokens. Two tokens are drawn without replacement. What is the probability the first is green and the second is yellow?

  4. From a standard deck of 52 cards, you draw two cards without replacement. What is the probability of drawing two Aces?

  5. A bin has 7 working light bulbs and 3 defective ones. You choose two bulbs without replacement. What is the probability of selecting at least one defective bulb?

  6. A committee of 3 students is chosen at random without replacement from a group of 6 girls and 4 boys. What is the probability that all 3 selected are girls?

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Probability of Draws Without Replacement

Answer key

  1. A bag contains 4 red balls and 6 green balls. You draw one ball, keep it, and draw a second. What is the probability that both balls are red?

    (4/10) * (3/9) = 12/90 = 2/15

  2. A drawer holds 8 black socks and 4 blue socks. You pull out two socks at random without replacement. What is the probability that both are black?

    (8/12) * (7/11) = 56/132 = 14/33

  3. A box contains 5 green tokens, 3 yellow tokens, and 2 white tokens. Two tokens are drawn without replacement. What is the probability the first is green and the second is yellow?

    (5/10) * (3/9) = 15/90 = 1/6

  4. From a standard deck of 52 cards, you draw two cards without replacement. What is the probability of drawing two Aces?

    (4/52) * (3/51) = 12/2652 = 1/221

  5. A bin has 7 working light bulbs and 3 defective ones. You choose two bulbs without replacement. What is the probability of selecting at least one defective bulb?

    1 - P(no defectives) = 1 - [(7/10) * (6/9)] = 1 - (42/90) = 48/90 = 8/15

  6. A committee of 3 students is chosen at random without replacement from a group of 6 girls and 4 boys. What is the probability that all 3 selected are girls?

    (6/10) * (5/9) * (4/8) = 120/720 = 1/6