Basic Probability and the Probability Scale

Can this result happen?

Check every number on the die.

Count the equal intervals between 0 and 1 before reading P.

Compare 0.8 with 0, one half and 1.

The denominator counts all counters.

List the even numbers on the die.

1. The even numbers are 2, 4 and 6: three favourable outcomes out of six. P(\text{even})=\frac36=\frac12.

Count sectors rather than different letters.

1. There are 8 sectors and 2 are B, so 8-2=6 are not B.
2. The probability is \frac68=\frac34.

The two probabilities add to 1.

1. P(\text{not late})=1-0.28=0.72.

All three probabilities add to 1.

1. The known probabilities total 0.35+0.40=0.75.
2. The missing probability is 1-0.75=0.25.

Check the limits of the probability scale.

1. A probability must be between 0 and 1 inclusive. The number 1.2 is greater than 1.

Update both the favourable count and the total after removing cards.

1. There are 20-2=18 cards left. The remaining multiples of 5 are 15 and 20.
2. P=\frac2{18}=\frac19.

Do not include 4: it is not greater than 4.

1. The numbers must meet both conditions. The even numbers greater than 4 are 6, 8 and 10.
2. There are three winning sectors out of ten, so the probability is \frac3{10}.

Work out how many 10p coins make £3, and how many 50p coins make £3.

1. £3 is 300p. Number of 10p coins: 300\div10=30. Number of 50p coins: 300\div50=6.
2. There are 30+6=36 coins, and 6 of them are 50p coins.
3. The probability of choosing a 50p coin is \frac{6}{36}=\frac{1}{6}.

Find each chance using winning tickets divided by the total. Compare the fractions.

Part (a)

1. Bag A has 3 winners among 8 tickets, so P(\text{win from A})=\frac38.

Part (b)

1. Bag B has 4 winners among 12 tickets, so P(\text{win from B})=\frac4{12}=\frac13.

Part (c)

1. \frac38=\frac9{24} and \frac13=\frac8{24}. Since \frac9{24}>\frac8{24}, Bag A gives the greater chance.

Find the given fraction of 32.

1. Red counters make up \frac38 of the total.
2. 32\div8\times3=4\times3=12 red counters.

The number of blue counters has not changed. What fraction of the new total is blue?

1. The new probability of blue is 1-\frac35=\frac25.
2. The 12 blue counters represent two of five equal parts. One part is 12\div2=6 counters, so the new total is 5\times6=30.
3. There were 8+12=20 counters initially. All the added counters were red, so 30-20=10 red counters were added.
4. Check: there are now 18 red counters out of 30, and \frac{18}{30}=\frac35.

Find the new total, then the new red count.

1. The new total is 5+15+10=30. The new red count is 30\div5\times2=12.
2. Red counters added: 12-5=7. Of the 10 added counters, 10-7=3 were blue.

The blue probability tells you what fraction the 9 counters represent.

1. The probability of blue is 1-\frac58=\frac38. Three equal parts represent 9 counters, so one part represents 9\div3=3.
2. Red is five parts: 5\times3=15 counters.

Split the remaining probability into parts in the ratio 2 to 1.

1. Win and loss together have probability 1-0.28=0.72.
2. A win takes two of three equal parts, so P(\text{win})=0.72\div3\times2=0.48.

An outcome that meets both conditions must only be counted once.

Part (a)

1. The numbers less than 5 are 1, 2, 3 and 4. The numbers greater than 2 are 3, 4, 5 and 6.
2. Together these include every number from 1 to 6. Count each once: the probability is \frac66=1.

Part (b)

1. The two events overlap at 3 and 4. Adding the two counts double-counts those results. A probability greater than 1 also signals an error.