A fair share of the rolls
Roll a fair dice 60 times. About how many sixes should you expect? The probability of a six is : one outcome in every six. So over 60 rolls, the sixes should get about one sixth of the rolls, their fair share. Every other face should get a sixth of the rolls too.
The 60 rolls in 6 equal shares of 10, one share for each face. The colored share is the sixes: about 10 of them.
The number of trials times the probability
One sixth of 60 is 60 ÷ 6 = 10. This number is called the expected count, and it is the number of trials × the probability. For the dice, , so the expected number of sixes in 60 rolls is about 10.
It works for any probability. A spinner lands on red with probability . In 80 spins, the expected number of reds is .
When the top number of the probability is more than 1, find one share first and then take as many shares as the top number says. A player scores a free throw with probability . In 40 free throws, a quarter of 40 is 10, so the expected number of baskets is 3 × 10 = 30.
The 40 free throws in 4 equal shares of 10. A probability of takes 3 of the shares: about 30 baskets.
Expected means about
The expected count is not a promise. Roll a dice 60 times and you might get 8 sixes, or 13, or exactly 10. Chance still swings from one run to the next.
What the expected count gives is where the results gather. Most runs of 60 rolls give a number of sixes somewhere near 10, and hardly any give 0 sixes or 30.
The colored mark is the expected count, 10 sixes. The other two marks are two real runs of 60 rolls, which gave 8 sixes and 13 sixes: both near 10, and neither exactly 10.
Multiply by the probability, not by its bottom number
The usual mistake is to multiply by the bottom number of the probability: 60 × 6 = 360 sixes, which is more sixes than there are rolls. A chance of means one roll in every 6, so divide: 60 ÷ 6 = 10.
An expected count can never be more than the number of trials, because a probability is never more than 1. That is a quick check on every answer.
Worked example: Visitors to a Science Museum Shop, Expected on Saturday and on Sunday
Question At a science museum, the probability that a visitor buys something in the shop is 310, on every day of the week. (a) On Saturday, 300 people visit the museum. How many of them would you expect to buy something? (b) On Sunday, the shop expects 72 buyers. How many visitors does the museum expect on Sunday?
1.A probability of 310 means about 3 buyers in every 10 visitors.
A probability of 310 means about 3 buyers in every 10 visitors. 2.(a) 300 visitors make 300 ÷ 10 = 30 groups of 10. Expect about 30 × 3 = 90 buyers. This is the same as 300 × 310 = 90.
(a) 300 visitors are 30 groups of 10: about 30 × 3 = 90 buyers. 3.On Sunday, 72 buyers make 72 ÷ 3 = 24 groups of 3, and each group of 3 buyers comes from a group of 10 visitors.
On Sunday, 72 buyers are 72 ÷ 3 = 24 groups of 3. 4.(b) Expect about 24 × 10 = 240 visitors. Check: 240 × 310 = 72.
(b) Each group of 3 buyers comes from 10 visitors: about 24 × 10 = 240 visitors.
Answer: (a) about 90 buyers; (b) about 240 visitors
Common mistakes
- Multiplying 72 by 310 in part (b) to get 21.6. The 72 buyers are only part of the visitors, so there must be more visitors than 72: divide by 3, then multiply by 10.
- Saying that exactly 90 people will buy something on Saturday. An expected count is about how many. On a real Saturday it might be 84 or 97.
Worked example: Free Throws in a Week of Practice, Checked Against the Coach's Claim
Question Jada practices free throws after school. She takes 20 free throws each day. She scores 13 baskets on Monday, 15 on Tuesday, 14 on Wednesday, 12 on Thursday and 16 on Friday. (a) What is the experimental probability that Jada scores a free throw? Give it as a decimal. (b) Her coach says Jada scores 3 out of every 4 free throws. How many baskets would the coach's claim predict for the week's free throws, and how many fewer did Jada really score?
1.Jada took 5 × 20 = 100 free throws. She scored 13 + 15 + 14 + 12 + 16 = 70 baskets.
She took 5 × 20 = 100 free throws and scored 70. 2.(a) The experimental probability is 70100 = 0.7.
(a) The experimental probability is 70100 = 0.7. 3.The coach's claim is 34. For 100 free throws it predicts 100 × 34 = 75 baskets.
The claim 34 predicts 100 × 34 = 75 baskets. 4.(b) Jada scored 75 − 70 = 5 fewer than the claim predicts. A gap of 5 in 100 shots is small, and chance alone often makes a gap that size, so this week does not show that the coach is wrong.
(b) She scored 75 − 70 = 5 fewer. A gap that small can come from chance alone.
Answer: (a) 0.7; (b) the claim predicts 75, and she scored 5 fewer
Common mistakes
- Dividing 70 by 5 days to get 14. That is the mean number of baskets in a day, not a probability. A probability is baskets out of shots: 70100.
- Saying the coach must be wrong because 70 is not 75. Results that depend on chance are rarely exactly what a probability predicts. A small gap over 100 shots is to be expected.