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Statistics 9-12 / Week 11 / Friday
5/6
Week 11 · Counting and the Multiplication Rule

Friday

Everyday draws and a review
// Volunteer shifts drawn two at a time
⏱ about 20 min

Friday: Everyday Draws and a Review

"Relay order," Comet says, fanning 4 pacer cards. "If we draw it by lot, what is the chance of exactly this order?"

"4! orders," Wren says. "24 of them, all equally likely. One order is 1/24."

"And two water-table helpers from 8 tokens," Comet says. "Chance at least one is blue?"

"Count none first," Wren says. "No blue is 15/28. So at least one is 13/28."

Nova projects the week in a row of cards: bag, rule, orders, groups, complement. "Would you like a hint? Each has a question it answers."

"What do you notice?" Wren asks. "Every draw by lot this week came down to counting equally likely things."

"Then Race Day can draw everything by lot," Comet says. "Lanes, shifts, pacers. Fair, and we can prove it."

Which tool answers which question?

QuestionToolAnswer
First red, then blue, without replacementP(A) × P(B given A)5/8 × 2/7 = 5/28
Exactly one relay order out of 241 ÷ P(4, 4)1/24
Cards A and B both in the morning group of 3C(4, 1) ÷ C(6, 3)4/20 = 1/5
Both tokens red, drawn at onceC(5, 2) ÷ C(8, 2)10/28 = 5/14
At least one blue in two tokens1 - P(no blue)1 - 15/28 = 13/28

Everyday words

"The second draw depends on the first" is the general multiplication rule in plain words. The bag shrank.

"How many ways can they line up?" is a permutation. "How many ways can we pick a team?" is a combination.

"At least one" means one or more. Count the chance of none and subtract from 1.

Review: the week in four lines

  • P(A and B) = P(A) × P(B given A) = P(B) × P(A given B).
  • P(n, k) counts orders; C(n, k) = P(n, k) ÷ k! counts groups.
  • Equally likely orders or groups: probability = favorable ÷ all.
  • P(at least one) = 1 - P(none).
FROM THE RUN LOG
  • Read the question.
  • Tap your answer.
The 4 pacers are placed in running order by lot. What is the probability of one particular order?
A bag holds 5 red, 2 blue, 1 gold tokens. Two are drawn without replacement. What is the probability the first is gold and the second is red?
From 6 volunteers, how many different pairs can take the afternoon shift?
A bag holds 8 tokens, 5 of them red tokens. 2 are drawn at once. What is the probability of at least one red token?
REASON IT OUT
  • Read the question.
  • Tap your answer.
Two tokens are drawn without replacement. Which statement is the careful one?
"How many ways can three volunteers be chosen for a shift?" Which count?
How many different pairs of tokens can be drawn at once from the 8?
With 4 pacers drawn into running order by lot, what is the probability of one particular order? Type the fraction.
WHY THIS EXERCISEOne favorable order over all equally likely orders. Counting the orders is the whole job.
StatementTrue or false?
4 × 3 × 2 × 1 = 24?
P(at least one blue) = P(blue first) + P(blue second).?
56 ÷ 2 = 28?
With replacement, P(red, red) = P(red) × P(red).?
WHY THIS EXERCISEThe review mixes the week: the rule, the two counts and the complement.
Try it
Design a draw by lot for a chore at home: one token per person, two drawn without replacement.
Find the probability that two named people are both drawn, by the rule and by combinations.
Draw the 4 pacer cards in one running order. Under them write 1/24 and how you counted the orders.

Fine week's work. Tomorrow is Track Day: a family draw without replacement.

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