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Statistics 9-12 / Week 11 / Monday
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Week 11 · Counting and the Multiplication Rule

Monday

Two draws without replacement
// Volunteer shifts drawn two at a time
⏱ about 20 min

Monday: Two Draws Without Replacement

Comet shakes the cloth bag. "8 shift tokens. 5 red for the water table, 2 blue for the markers, 1 gold for the timer."

"Draw two for the first pair of volunteers," Wren says. "Chance the first is red and the second is blue?"

"Red first is 5 out of 8," Comet says, drawing a red. "Now blue."

"Stop," Wren says. "What do you notice about the bag now? Only 7 tokens are left."

"So blue is 2 out of 7, not 8," Comet says. "The first draw changed the second."

Nova projects the two fractions side by side. "Would you like a hint? Multiply them."

"5/8 × 2/7 = 5/28," Wren says. "That is the general multiplication rule."

"Then let us count every pair of draws," Comet says, "and see if the rule holds."

Name cards in rows on the volunteer board; Wren counts arrangements while Comet pulls two tokens from the bag.

Without replacement

Draw a token and keep it out of the bag. The second draw comes from a smaller bag. That is drawing without replacement.

The probability of the second draw depends on what the first draw took. We write it as P(B given A).

Red first: P(A) = 5/8. Blue second, given red first: P(B given A) = 2/7, from the 7 tokens left.

The general multiplication rule: P(A and B) = P(A) × P(B given A). Read it as "first A, then B knowing A happened."

Every number here is the crew's own bag of tokens from Nova's Run Log, not a fact about any real draw.

A solved problem to study

  1. The bag holds 5 red, 2 blue and 1 gold, 8 tokens in all.
  2. P(red first) = 5/8 = 5/8.
  3. After a red is taken, 7 tokens remain and 2 are blue. P(blue second given red first) = 2/7.
  4. P(red then blue) = 5/8 × 2/7 = 5/28.
  5. Check by counting: 8 × 7 = 56 equally likely ordered pairs. Red then blue pairs: 5 × 2 = 10. So 10/56 = 5/28.
  6. Two reds in a row: 5/8 × 4/7 = 5/14. After one red, only 4 reds remain.

Notice the second fraction. Its bottom number drops by one, and its top number drops only if the same color was taken.

DrawTokens left before itProbability
red first85/8
blue second, after a red72/7
red second, after a red74/7
gold second, after a red71/7
TWO DRAWS WITHOUT REPLACEMENT
  • Read the question.
  • Tap your answer.
A bag holds 5 red, 2 blue, 1 gold tokens. Two are drawn without replacement. What is the probability the first is red and the second is blue?
A bag holds 5 red, 2 blue, 1 gold tokens. Two are drawn without replacement. What is the probability the first is red and the second is red?
A bag holds 5 red, 2 blue, 1 gold tokens. Two are drawn without replacement. What is the probability the first is blue and the second is gold?
A second bag holds 4 red and 2 blue tokens. Two are drawn without replacement. What is the probability both are blue?
A red token is drawn and kept out. How many tokens are left in the bag? Type the number.
WHY THIS EXERCISEThe smaller bag is the whole idea of "given". The second fraction's bottom number is the bag after the first draw.
StatementTrue or false?
Without replacement, the second draw comes from a smaller bag.?
5 × 2 = 10?
P(A and B) = P(A) + P(B given A).?
8 × 7 = 56?
After one red is drawn, the chance of another red goes down.?
WHY THIS EXERCISEMultiplying a probability by a conditional probability is how two steps become one event.
Try it
Put 8 paper scraps in a cup, 5 marked R, 2 marked B, 1 marked G.
Draw two without putting the first back. Do it ten times and tally how often you get R then B.
Draw the bag before the first draw and after it. Write the fraction for red under the first and blue under the second.

Strong start. Tomorrow the crew counts shift orders and shift groups, and the two counts turn into probabilities.