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Statistics 9-12 / Week 11 / Thursday
4/6
Week 11 · Counting and the Multiplication Rule

Thursday

Why the rule works, and at least one
// Volunteer shifts drawn two at a time
⏱ about 20 min

Thursday: Why the Rule Works, and At Least One

"Prove it," Comet says. "Why does P(A) × P(B given A) give the right answer every time?"

Wren rules a grid: 8 rows for the first token, 7 columns for the second. "56 boxes, all equally likely."

"Red first fills 5 rows," he says. "5 × 7 = 35 boxes. P(A) = 35/56."

"Inside those rows, blue second is 10 boxes," Comet says. "So P(B given A) = 10/35."

"Multiply," Wren says. "35/56 × 10/35. The 35 cancels. 10/56. The same boxes."

Nova glows. "Would you like a hint? The rule also runs the other way, second draw first."

"What do you notice?" Wren asks. "P(B) × P(A given B) is 1/4 × 5/7. Also 5/28."

Why P(A and B) = P(A) × P(B given A)

Monday's rule, proved by counting the ordered pairs of draws. Read it first, then put it in order.

  1. The bag has 8 tokens, so two draws make 8 × 7 = 56 ordered pairs, all equally likely.
  2. Pairs with a red first: 5 × 7 = 35. So P(A) = 35/56.
  3. Among those 35 pairs, the ones with a blue second: 5 × 2 = 10. So P(B given A) = 10/35.
  4. Pairs with red first and blue second are those same 10, out of all 56. So P(A and B) = 10/56.
  5. Multiply: 35/56 × 10/35 = 10/56. The 35 cancels.
  6. In general: (A pairs ÷ all) × (A and B pairs ÷ A pairs) = A and B pairs ÷ all. That is the rule.
THE PROOF, IN ORDER
  • ?Two draws make 56 equally likely ordered pairs
  • ?35/56 × 10/35 = 10/56, the shared 35 cancels
  • ?10 of those have a blue second, so P(B given A) = 10/35
  • ?35 pairs have a red first, so P(A) = 35/56
  • ?Those 10 pairs are the A and B pairs, so P(A and B) = 10/56
  • ?So P(A and B) = P(A) × P(B given A)
WHY THIS EXERCISEThe rule is one cancelling fraction. Counting the pairs is what makes it true.

The rule runs both ways

P(A and B) = P(B) × P(A given B) too. Blue second: 2 × 7 = 14 pairs, so P(B) = 1/4.

Among those 14, red first is 10: P(A given B) = 5/7. And 1/4 × 5/7 = 5/28.

With replacement the draws are independent, and P(red, red) would be 5/8 × 5/8 = 25/64, not 5/14.

At least one, with combinations

Draw two tokens at once. All C(8, 2) = 28 pairs are equally likely.

No blue: both tokens come from the 6 non-blue tokens, C(6, 2) = 15 pairs. P(no blue) = 15/28.

At least one blue is everything else: 1 - 15/28 = 13/28.

"At least one" is the complement of "none". Counting none is almost always easier.

USE THE RULE AND THE COUNTS
  • Read the question.
  • Tap your answer.
P(red first) = 5/8 and P(blue second given red first) = 2/7. What is P(red first and blue second)?
P(blue second) = 1/4 and P(red first given blue second) = 5/7. What is P(blue second and red first)?
A bag holds 8 tokens, 2 of them blue tokens. 2 are drawn at once. What is the probability of at least one blue token?
A bag holds 8 tokens, 2 of them blue tokens. 2 are drawn at once. What is the probability all 2 are blue tokens?
WHICH CONCLUSION IS JUSTIFIED?
  • Read the question.
  • Tap your answer.
With replacement, P(red, red) is 25/64. Without, it is 5/14. Why does it drop?
Which is the easier way to find P(at least one blue in two draws)?
P(B given A) is called a this probability. Type one word.
"At least one" is the this of "none". Type one word.
StatementTrue or false?
35/56 × 10/35 = 5/28?
P(A) × P(B given A) and P(B) × P(A given B) give the same P(A and B).?
1 - 15/28 = 13/28?
With replacement, the second draw depends on the first.?
WHY THIS EXERCISEThe proof, the two orders and the complement are the whole of this week's rule.
Try it
On graph paper, draw the 8 by 7 grid of ordered pairs. Shade the red-first rows and circle the blue-second boxes.
Count the circled boxes inside the shade. Check them against the rule.

Sharp thinking. Tomorrow the rule and the counts meet everyday questions, and the week gets its review.

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