Grade 11 · lesson 4 of 4

Track Two Draws Without Replacement

Multiply a first-draw probability by the correctly updated second-draw probability.

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Supplied materials

  • Paper or a note app.
  • Fictional uniform box: 3 red cards and 2 blue cards. Draw two cards in order without replacing the first.

Learn the idea

Without replacement, the first card is absent during the second draw. For red then blue, the first red probability is 3/5. After that red draw, four cards remain and two are blue, so the next blue probability is 2/4.

Multiply along this specified path: P(red then blue) = P(red first) × P(blue second given red first). This is a conditional multiplication rule. You cannot keep the original denominator five for the second draw. Replacement would produce a different model.

  1. Write the starting counts as 3 red, 2 blue, total 5.
  2. On the red-first branch, update to 2 red, 2 blue, total 4.
  3. Multiply 3/5 by 2/4, then update both counts for the new two-red question.
Worked example

See the method in action

P(red then blue) = (3/5)(2/4) = 6/20 = 3/10. The second probability is conditioned on red having been removed.

Your turn

Check your reasoning

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A new box has 4 red and 2 blue cards. Without replacement, what is the chance that both ordered draws are red?

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See the answer and explanation

2/5

First red is 4/6. After it is removed, three red cards remain among five, so the chance is (4/6)(3/5) = 12/30 = 2/5.

Take it further

Replace the first card before the second draw in that 4-red, 2-blue box. Both red now has probability (4/6)(4/6) = 4/9, because the second sample space is reset.

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