Faded examples · Two picks without replacement: P(both the same)

Non-calculator

Each worked example shows a little less than the one before – you complete the faded (blank) steps yourself, following the same four steps every time. The items are taken WITHOUT replacement (not put back). In each example also answer the check question. No calculator.

①Example 1fully worked: read it through

A bag has $5$ red and $3$ blue counters. Two are taken without replacement. Find P(both red).

1First-pick probability

first pick: P(red) $=\dfrac{5}{8}$

2Second-pick probability (one fewer)

not replaced, so $4$ red of $7$ left: $\dfrac{4}{7}$

3Multiply along the branch

$\dfrac{5}{8}\times\dfrac{4}{7}=\dfrac{20}{56}$

4Simplify the answer

simplify: $\dfrac{20}{56}=\dfrac{5}{14}$

Check · Why is the second probability $\dfrac{4}{7}$ and not $\dfrac{5}{8}$?

A the first red is not replaced, so one fewer red and one fewer counter remainB because $5-1=4$ and $8-1=7$ always, for any questionC because the two picks must use different denominatorsD it is a mistake – it should be $\dfrac{5}{8}$

②Example 2you finish the last 1 step

A box has $4$ green and $2$ yellow beads. Two are taken without replacement. Find P(both green).

1First-pick probability

first pick: P(green) $=\dfrac{4}{6}$

2Second-pick probability (one fewer)

not replaced, so $3$ green of $5$ left: $\dfrac{3}{5}$

3Multiply along the branch

$\dfrac{4}{6}\times\dfrac{3}{5}=\dfrac{12}{30}$

4Simplify the answer

simplify: $\dfrac{12}{30}=\dfrac{2}{5}$

Check · A student worked out $\dfrac{4}{6}\times\dfrac{4}{6}$. What did they forget?

A to add the two routes togetherB that a bead is removed, so the second pick has different numbersC to change the first fraction as wellD nothing – that is correct

③Example 3you finish the last 2 steps

A jar has $6$ red and $4$ blue sweets. Two are taken without replacement. Find P(both red).

1First-pick probability

first pick: P(red) $=\dfrac{6}{10}$

2Second-pick probability (one fewer)

not replaced, so $5$ red of $9$ left: $\dfrac{5}{9}$

3Multiply along the branch

$\dfrac{6}{10}\times\dfrac{5}{9}=\dfrac{30}{90}$

4Simplify the answer

simplify: $\dfrac{30}{90}=\dfrac{1}{3}$

Check · To combine the two branches of the "red then red" route, you...

A add the probabilitiesB subtract the probabilitiesC multiply the probabilities along the branchD divide the probabilities

④Example 4your turn: every step

A box has $3$ white and $5$ black beads. Two are taken without replacement. Find P(both white).

1First-pick probability

first pick: P(white) $=\dfrac{3}{8}$

2Second-pick probability (one fewer)

not replaced, so $2$ white of $7$ left: $\dfrac{2}{7}$

3Multiply along the branch

$\dfrac{3}{8}\times\dfrac{2}{7}=\dfrac{6}{56}$

4Simplify the answer

simplify: $\dfrac{6}{56}=\dfrac{3}{28}$

Check · How would you find P(at least one white) for this bag instead?

A work out $1-$ P(both black)B work out P(both white) $+1$C multiply the P(both white) answer by $2$D you cannot find it from a tree

Answers · Probability Trees

Faded examples · Two picks without replacement: P(both the same)

① Example 1 first pick: P(red) $=\dfrac{5}{8}$→not replaced, so $4$ red of $7$ left: $\dfrac{4}{7}$→$\dfrac{5}{8}\times\dfrac{4}{7}=\dfrac{20}{56}$→simplify: $\dfrac{20}{56}=\dfrac{5}{14}$ $\dfrac{5}{14}$

Check: A: the first red is not replaced, so one fewer red and one fewer counter remain

② Example 2 first pick: P(green) $=\dfrac{4}{6}$→not replaced, so $3$ green of $5$ left: $\dfrac{3}{5}$→$\dfrac{4}{6}\times\dfrac{3}{5}=\dfrac{12}{30}$→simplify: $\dfrac{12}{30}=\dfrac{2}{5}$ $\dfrac{2}{5}$

Check: B: that a bead is removed, so the second pick has different numbers

③ Example 3 first pick: P(red) $=\dfrac{6}{10}$→not replaced, so $5$ red of $9$ left: $\dfrac{5}{9}$→$\dfrac{6}{10}\times\dfrac{5}{9}=\dfrac{30}{90}$→simplify: $\dfrac{30}{90}=\dfrac{1}{3}$ $\dfrac{1}{3}$

Check: C: multiply the probabilities along the branch

④ Example 4 first pick: P(white) $=\dfrac{3}{8}$→not replaced, so $2$ white of $7$ left: $\dfrac{2}{7}$→$\dfrac{3}{8}\times\dfrac{2}{7}=\dfrac{6}{56}$→simplify: $\dfrac{6}{56}=\dfrac{3}{28}$ $\dfrac{3}{28}$

Check: A: work out $1-$ P(both black)