1fully worked

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

1first pick: P(red) $=\dfrac{5}{8}$First-pick probability

2not replaced, so $4$ red of $7$ left: $\dfrac{4}{7}$Second-pick probability (one fewer)

3$\dfrac{5}{8}\times\dfrac{4}{7}=\dfrac{20}{56}$Multiply along the branch

4simplify: $\dfrac{20}{56}=\dfrac{5}{14}$Simplify the answer

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 $\dfrac{5}{8}$

2you finish it

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

1first pick: P(green) $=\dfrac{4}{6}$First-pick probability

2not replaced, so $3$ green of $5$ left: $\dfrac{3}{5}$Second-pick probability (one fewer)

3$\dfrac{4}{6}\times\dfrac{3}{5}=\dfrac{12}{30}$Multiply along the branch

4Simplify the answer

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

3you finish it

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

1first pick: P(red) $=\dfrac{6}{10}$First-pick probability

2not replaced, so $5$ red of $9$ left: $\dfrac{5}{9}$Second-pick probability (one fewer)

3Multiply along the branch

4Simplify the answer

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

4you do it all

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

1First-pick probability

2Second-pick probability (one fewer)

3Multiply along the branch

4Simplify the answer

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