Probability · Ordinary Level
With & Without Replacement
Ordinary Level · Class 5 · Tap NEXT to begin
Section 1 of 3
With Replacement
Put it back
If the first item is replaced, the second draw sees the same bag — the fractions don’t change.
A bag has $3$ blue and $5$ green marbles ($8$ total). One is taken, its colour noted, put back, then a second is taken.
Worked example — both, and only the first
Find (i) $P(\text{both blue})$; (ii) $P(\text{both green})$; (iii) $P(\text{only the first is green})$.
$P(\text{blue}) = \tfrac{3}{8},\ P(\text{green}) = \tfrac{5}{8}$
(i) $\tfrac{3}{8} \times \tfrac{3}{8} = \tfrac{9}{64}$
(ii) $\tfrac{5}{8} \times \tfrac{5}{8} = \tfrac{25}{64}$
(iii) green and blue $= \tfrac{5}{8} \times \tfrac{3}{8} = \tfrac{15}{64}$
$\tfrac{9}{64},\ \tfrac{25}{64},\ \tfrac{15}{64}$
Section 2 of 3
Without Replacement
Keep it out
If the first is not replaced, the second draw has one fewer in total and one fewer of the colour taken.
A bag has $7$ green and $3$ red marbles ($10$ total). One is taken and not replaced, then a second.
Worked example — both green, both red, same, only one
Find (i) $P(\text{both green})$; (ii) $P(\text{both red})$; (iii) $P(\text{same colour})$; (iv) $P(\text{only one green})$.
(i) $\tfrac{7}{10} \times \tfrac{6}{9} = \tfrac{7}{15}$
(ii) $\tfrac{3}{10} \times \tfrac{2}{9} = \tfrac{1}{15}$
(iii) both green or both red $= \tfrac{7}{15} + \tfrac{1}{15} = \tfrac{8}{15}$
(iv) GR or RG $= \tfrac{7}{10}\times\tfrac{3}{9} + \tfrac{3}{10}\times\tfrac{7}{9} = \tfrac{7}{30} + \tfrac{7}{30} = \tfrac{7}{15}$
$\tfrac{7}{15},\ \tfrac{1}{15},\ \tfrac{8}{15},\ \tfrac{7}{15}$
Section 3 of 3
Another Without-Replacement Bag
A bag has $5$ blue and $9$ pink marbles ($14$ total). One is taken and not replaced, then a second.
Worked example — both, and the second
Find (i) $P(\text{both blue})$; (ii) $P(\text{both pink})$.
(i) $\tfrac{5}{14} \times \tfrac{4}{13} = \tfrac{10}{91}$
(ii) $\tfrac{9}{14} \times \tfrac{8}{13} = \tfrac{36}{91}$
$\tfrac{10}{91},\ \tfrac{36}{91}$
You try
For that bag ($5$ blue, $9$ pink, no replacement), find $P(\text{the second one is pink})$.
Pink then pink, OR blue then pink — add them.
pink and pink $= \tfrac{9}{14} \times \tfrac{8}{13} = \tfrac{36}{91}$
blue and pink $= \tfrac{5}{14} \times \tfrac{9}{13} = \tfrac{45}{182}$
$\tfrac{36}{91} + \tfrac{45}{182} = \tfrac{117}{182} = \tfrac{9}{14}$
$\tfrac{9}{14}$
That’s Class 5.
Two draws with replacement and without replacement. Class 6: sample-space grids for two events.