Tree diagrams: without replacement

The Maths Habit

  1. A spinner has P(red) = 0.3. It is spun twice. Find P(red both times).

    0.09

  2. A bag holds 4 green and 3 white counters. One is taken and replaced, then another is taken. Find P(both white).

    949

  3. P(Mia’s train is on time) = 0.9. Find P(on time on exactly one of two days).

    0.18

A bag holds 5 red and 5 blue sweets.
You take a red one and eat it.
Is the next sweet more likely to be red or blue?

Blue: 4 red and 5 blue are left

Without replacement
The first counter is not put back, so there is one fewer for the second pick.

Find P(red) for the second counter

3 red and 2 blue. A red has been taken 24

Find P(blue) for the second counter

3 red and 2 blue. A red has been taken 24

Multiply or add?

Red AND then red Multiply

A bag holds 4 red and 2 blue counters. Two are taken without replacement. Find the probability that both are red.

first countersecond counterRRRRBRBBRBRBBB4/63/54/6 × 3/5 = 12/302/54/6 × 2/5 = 8/302/64/52/6 × 4/5 = 8/301/52/6 × 1/5 = 2/30
Answer25

A bag holds 5 green and 2 yellow counters. Two are taken without replacement. Find the probability that both are green.

first countersecond counterG5/7GGGYGYY2/7GYGYYY4/65/7 × 4/6 = 20/422/65/7 × 2/6 = 10/425/62/7 × 5/6 = 10/421/62/7 × 1/6 = 2/42
Answer1021

A bag holds 4 red and 2 blue counters. Two are taken without replacement. Find the probability that both are red.

first countersecond counterRRRRBRBBRBRBBB4/63/54/6 × 3/5 = 12/302/54/6 × 2/5 = 8/302/64/52/6 × 4/5 = 8/301/52/6 × 1/5 = 2/30
Answer25

A bag holds 3 red and 3 blue counters. Two are taken without replacement. Find the probability that both are blue.

first countersecond counterR3/6RRRBRBB3/6RBRBBB2/53/6 × 2/5 = 6/303/53/6 × 3/5 = 9/303/53/6 × 3/5 = 9/302/53/6 × 2/5 = 6/30
Answer15

A bag holds 4 red and 2 blue counters. Two are taken without replacement. Find the probability that both are red.

first countersecond counterRRRRBRBBRBRBBB4/63/54/6 × 3/5 = 12/302/54/6 × 2/5 = 8/302/64/52/6 × 4/5 = 8/301/52/6 × 1/5 = 2/30
Answer25

A bag holds 5 red and 4 blue counters.
Two are taken without replacement.
Find the probability that both are red.

first countersecond counterRRRRBRBBRBRBBB5/94/85/9 × 4/8 = 20/724/85/9 × 4/8 = 20/724/95/84/9 × 5/8 = 20/723/84/9 × 3/8 = 12/72
Answer518

A bag holds 4 red and 2 blue counters. Two are taken without replacement. Find the probability that both are red.

first countersecond counterRRRRBRBBRBRBBB4/63/54/6 × 3/5 = 12/302/54/6 × 2/5 = 8/302/64/52/6 × 4/5 = 8/301/52/6 × 1/5 = 2/30
Answer25

A bag holds 4 red and 3 blue counters.
Two are taken without replacement.
Find P(one of each colour).

first countersecond counterRRRRBRBBRBRBBB4/73/64/7 × 3/6 = 12/423/64/7 × 3/6 = 12/423/74/63/7 × 4/6 = 12/422/63/7 × 2/6 = 6/42
Answer2442 = 47

A bag holds 4 red and 2 blue counters. Two are taken without replacement. Find the probability that both are red.

first countersecond counterRRRRBRBBRBRBBB4/63/54/6 × 3/5 = 12/302/54/6 × 2/5 = 8/302/64/52/6 × 4/5 = 8/301/52/6 × 1/5 = 2/30
Answer25

A bag holds 2 green and 5 yellow counters.
Two are taken without replacement.
Find P(both the same colour).

first countersecond counterGGGGYGYYGYGYYY2/71/62/7 × 1/6 = 2/425/62/7 × 5/6 = 10/425/72/65/7 × 2/6 = 10/424/65/7 × 4/6 = 20/42
Answer2242 = 1121

Tree diagrams: without replacement worksheet, page 1

Answer sheet

  1. A box holds 7 milk and 3 dark chocolates. Sam eats two at random. Find the probability that at least one is dark.

    815

  2. A bag holds 4 red counters and some blue ones. Two are taken without replacement. P(both red) = 25. How many blue counters are there?

    2

  3. A tin holds 5 plain and 4 iced biscuits. Two are taken at random without replacement. Find the probability that one of each type is taken.

    59

  1. Two counters are taken without replacement from a bag of 5 red and 3 blue. Fill in the tree diagram.

    47, 37; 57, 27
    58RedRedBlue38BlueRedBlue

  2. Find the probability that both counters are red.

    2056 = 514

  3. Find the probability of one counter of each colour.

    3056 = 1528
Exit tickets to print