Finding an Unknown from a Given Probability

Working backwards from a probability: the total number of objects, an unknown number of balls, balls added or removed, the angle of a sector, and whether a stated probability is possible.

What You Should Already Know

From lessons 13.2 and 13.5: $P(E) =$ number of favourable outcomes $\div$ total number of possible outcomes, and $P(\text{sector}) =$ angle $\div 360^\circ$ .
Solving an equation in which a fraction equals a fraction (chapters 1 and 4): multiply both sides by both denominators, or cross-multiply. If $\frac{3}{x} = \frac{1}{4}$ , then $3 \times 4 = 1 \times x$ , so $x = 12$ .

The Total, or the Number Favourable

So far we have been given the objects and asked for a probability. In this lesson the probability is given, and something about the objects is unknown. The formula still links them; we write it as an equation and solve it.

Three forms of one formula

For equally likely outcomes, $P = \dfrac{\text{favourable}}{\text{total}}\text{,}$ so

favourable $= P \times$ total, and total $=$ favourable $\div P$ .

Every number of objects is a whole number. Check that your answer is, and that it gives back the probability you were given.

Example 1: Finding the total

A bag contains 8 green balls and some balls of other colours. A ball is drawn at random, and the probability that it is green is $\frac{1}{4}$ . How many balls are in the bag?

Solution

Let the total be $T$ . Then $\frac{8}{T} = \frac{1}{4}$ .

Cross-multiply: $8 \times 4 = 1 \times T$ , so $T = 32$ .

Check: $\frac{8}{32} = \frac{1}{4}$ .

There are 32 balls in the bag.

Example 2: Finding the number favourable

A pupil is chosen at random from a class of 40. The probability that the pupil wears glasses is $\frac{3}{8}$ . How many pupils in the class wear glasses?

Solution

Favourable $= P \times$ total $= \frac{3}{8} \times 40 = 15$ .

Check: $\frac{15}{40} = \frac{3}{8}$ .

15 pupils wear glasses.

An Unknown Number in a Bag

Example 3: How many blue balls?

A bag contains 6 red balls and $x$ blue balls. A ball is drawn at random, and $P(\text{red}) = \frac{2}{5}$ . Find $x$ .

Solution

Total number of balls $= 6 + x$ , so $P(\text{red}) = \frac{6}{6 + x}\text{.}$

$\frac{6}{6 + x} = \frac{2}{5}$

Cross-multiply: $5 \times 6 = 2(6 + x)$ , so $30 = 12 + 2x$ .

$2x = 18$ , so $x = 9$ .

Check: $\frac{6}{6 + 9} = \frac{6}{15} = \frac{2}{5}\text{.}$

$x = 9$ : there are 9 blue balls.

Brackets matter

Cross-multiplying $\frac{6}{6 + x} = \frac{2}{5}$ gives $30 = 2(6 + x)$ , not $30 = 12 + x$ . The 2 multiplies the whole of $6 + x$ .

Balls Added or Removed

When balls are added or taken out, both the number favourable and the total may change. Write the new counts first, then the new probability.

Example 4: Adding red balls

A bag contains 4 red balls and 8 blue balls. (a) 3 more red balls are put into the bag. Find the new probability of drawing a red ball. (b) Instead, how many red balls must be added to make the probability of drawing a red ball $\frac{1}{2}$ ?

Solution

(a) Red: $4 + 3 = 7$ . Total: $12 + 3 = 15$ . $P(\text{red}) = \frac{7}{15}$ .

(b) Let $k$ red balls be added. Red: $4 + k$ . Total: $12 + k$ .

$\frac{4 + k}{12 + k} = \frac{1}{2}\text{,}$ so $2(4 + k) = 12 + k$ , $8 + 2k = 12 + k$ , $k = 4$ .

Check: $\frac{8}{16} = \frac{1}{2}$ : 8 red and 8 blue, an even chance.

(a) $\frac{7}{15}$ ; (b) 4 red balls.

Example 5: Removing blue balls

A bag contains 5 red balls and 15 blue balls. Some blue balls are taken out, and now the probability of drawing a red ball is $\frac{1}{3}$ . How many blue balls were taken out?

Solution

Let $k$ blue balls be taken out. Red stays 5; the total becomes $20 - k$ .

$\frac{5}{20 - k} = \frac{1}{3}$ , so $5 \times 3 = 20 - k$ , $15 = 20 - k$ , $k = 5$ .

Check: 5 red and 10 blue are left, and $\frac{5}{15} = \frac{1}{3}$ .

5 blue balls were taken out.

Spinners and Cards

Example 6: The angle of a sector

A spinner has red, blue and green sectors. The green sector is $100^\circ$ and the probability of landing on red is $\frac{1}{4}$ . Find (a) the angle of the red sector; (b) the angle of the blue sector and the probability of landing on blue.

Solution

(a) $\frac{\text{angle}}{360^\circ} = \frac{1}{4}\text{,}$ so the red angle $= \frac{1}{4} \times 360^\circ = 90^\circ\text{.}$

(b) Blue angle $= 360^\circ - 90^\circ - 100^\circ = 170^\circ\text{.}$

$P(\text{blue}) = \frac{170}{360} = \frac{17}{36}$ (dividing both by 10).

Check: $P(\text{green}) = \frac{100}{360} = \frac{10}{36}\text{,}$ and $\frac{9}{36} + \frac{17}{36} + \frac{10}{36} = 1\text{.}$

(a) $90^\circ$ ; (b) $170^\circ$ , $P(\text{blue}) = \frac{17}{36}\text{.}$

Example 7: Part of a pack

Ravi takes all 13 hearts and some of the spades from a pack of playing cards and shuffles them. A card is drawn at random from his pile, and the probability that it is a spade is $\frac{5}{18}$ . How many spades are in the pile?

Solution

Let there be $s$ spades. The pile has $13 + s$ cards.

$\frac{s}{13 + s} = \frac{5}{18}$ , so $18s = 5(13 + s) = 65 + 5s\text{.}$

$13s = 65$ , so $s = 5$ .

Check: $\frac{5}{18}$ with $13 + 5 = 18$ cards, and 5 is no more than the 13 spades of a pack.

There are 5 spades in the pile.

Is the Probability Possible?

A probability given for a set of objects must give a whole number of favourable objects, from 0 to the total.

Example 8: A bag of 10 balls

A bag contains 10 balls, some of them red. Could the probability of drawing a red ball be (a) $\frac{3}{7}$ ; (b) $\frac{2}{5}$ ; (c) $\frac{11}{10}$ ? Explain.

Solution

(a) Red balls $= \frac{3}{7} \times 10 = \frac{30}{7} = 4\frac{2}{7}\text{:}$ not a whole number. Not possible.

(b) Red balls $= \frac{2}{5} \times 10 = 4$ : possible, with 4 red balls.

(a) no; (b) yes, 4 red balls; (c) no.

Common Mistakes

Forgetting that the total includes the unknown: with 6 red and $x$ blue balls, $P(\text{red}) = \frac{6}{6 + x}\text{,}$ not $\frac{6}{x}$ .
Dropping the brackets when cross-multiplying: $2(6 + x) = 12 + 2x$ , not $12 + x$ .
Changing only the number favourable when balls are added: adding 3 red balls adds 3 to the total as well.
Giving an answer that is not a whole number of objects without noticing that something is wrong.
Answering with the unknown when the question asks for something else, such as the total or the new probability.

Practice

Try each question first, then open the solution to check your working. Check every answer by putting it back into the probability.

Question 1

A box contains 12 yellow counters and some counters of other colours. The probability of taking a yellow counter at random is $\frac{3}{10}$ . How many counters are in the box?

ShowHide solution

$\frac{12}{T} = \frac{3}{10}$ , so $3T = 120$ and $T = 40$ .

Check: $\frac{12}{40} = \frac{3}{10}$ .

40 counters.

Question 2

At a school carnival, 60 lucky-draw tickets are in a box, and some of them win a prize. The probability that a ticket drawn at random wins a prize is $\frac{1}{12}$ . How many tickets win a prize?

ShowHide solution

$\frac{1}{12} \times 60 = 5$ .

Check: $\frac{5}{60} = \frac{1}{12}$ .

5 tickets.

Question 3

A pencil case holds 5 blue pens and $n$ black pens. A pen is taken at random, and $P(\text{black}) = \frac{3}{4}$ . Find $n$ .

ShowHide solution

$\frac{n}{5 + n} = \frac{3}{4}$ , so $4n = 3(5 + n) = 15 + 3n\text{.}$

$n = 15$ .

Check: $\frac{15}{20} = \frac{3}{4}$ .

$n = 15$ .

Question 4

A bag contains 10 red marbles and $x$ white marbles. The probability of drawing a white marble at random is $\frac{1}{3}$ . Find $x$ , and the probability of drawing a red marble.

ShowHide solution

$\frac{x}{10 + x} = \frac{1}{3}$ , so $3x = 10 + x$ , $2x = 10$ , $x = 5$ .

$P(\text{red}) = \frac{10}{15} = \frac{2}{3}$ (or $1 - \frac{1}{3}$ ).

$x = 5$ ; $P(\text{red}) = \frac{2}{3}$ .

Question 5

A bag contains 3 green balls and 9 yellow balls. (a) 2 more green balls are added. Find the new probability of drawing a green ball. (b) Instead, how many green balls must be added to make the probability of drawing a green ball $\frac{1}{2}$ ?

ShowHide solution

(a) Green 5, total 14: $\frac{5}{14}$ .

(b) Let $k$ green balls be added: green $3 + k$ , total $12 + k$ .

$\frac{3 + k}{12 + k} = \frac{1}{2}\text{,}$ so $2(3 + k) = 12 + k$ , $6 + 2k = 12 + k$ , $k = 6$ .

Check: $\frac{9}{18} = \frac{1}{2}$ .

(a) $\frac{5}{14}$ ; (b) 6 green balls.

Question 6

A bag contains 6 red balls and 10 blue balls. Some red balls are taken out, and now the probability of drawing a red ball is $\frac{1}{6}$ . How many red balls were taken out?

ShowHide solution

Let $k$ red balls be taken out: red $6 - k$ , total $16 - k$ .

$\frac{6 - k}{16 - k} = \frac{1}{6}\text{,}$ so $6(6 - k) = 16 - k$ , $36 - 6k = 16 - k$ , $20 = 5k$ , $k = 4$ .

Check: 2 red and 10 blue are left, and $\frac{2}{12} = \frac{1}{6}$ .

4 red balls.

Question 7

A spinner has blue, red and yellow sectors. $P(\text{blue}) = \frac{3}{8}$ and $P(\text{red}) = \frac{1}{6}$ . Find the angle of each sector and $P(\text{yellow})$ .

ShowHide solution

Blue: $\frac{3}{8} \times 360^\circ = 135^\circ\text{.}$ Red: $\frac{1}{6} \times 360^\circ = 60^\circ\text{.}$

Yellow: $360^\circ - 135^\circ - 60^\circ = 165^\circ\text{,}$ and $P(\text{yellow}) = \frac{165}{360} = \frac{11}{24}$ (dividing both by 15).

Check: $\frac{9}{24} + \frac{4}{24} + \frac{11}{24} = 1\text{.}$

Blue $135^\circ$ , red $60^\circ$ , yellow $165^\circ$ ; $P(\text{yellow}) = \frac{11}{24}\text{.}$

Question 8

The red sector of a spinner has an angle of $x^\circ$ , and the probability of landing on red is $\frac{2}{9}$ . Find $x$ .

ShowHide solution

$\frac{x}{360} = \frac{2}{9}$ , so $9x = 720$ and $x = 80$ .

$x = 80$ .

Question 9

Mei Ling takes all 13 clubs and some of the diamonds from a pack of playing cards. A card is drawn at random from her pile, and the probability that it is a diamond is $\frac{7}{20}$ . How many diamonds are in her pile?

ShowHide solution

Let there be $d$ diamonds: $\frac{d}{13 + d} = \frac{7}{20}$ .

$20d = 7(13 + d) = 91 + 7d\text{,}$ so $13d = 91$ and $d = 7$ .

Check: $\frac{7}{20}$ with $13 + 7 = 20$ cards.

7 diamonds.

Question 10

A box contains 12 counters, some of them red. Could the probability of taking a red counter be (a) $\frac{5}{8}$ ; (b) $\frac{5}{6}$ ; (c) $\frac{3}{4}$ ? Explain.

ShowHide solution

(a) $\frac{5}{8} \times 12 = 7.5$ : not a whole number, so not possible.

(b) $\frac{5}{6} \times 12 = 10$ : possible, 10 red counters.

(a) no; (b) yes, 10; (c) yes, 9.

Question 11

A class of 36 pupils is to choose a pupil at random to lead a class activity. The probability that a girl is chosen is $\frac{4}{9}$ . (a) How many girls are in the class? (b) 4 more girls join the class before the choice is made. Find the new probability that a girl is chosen.

ShowHide solution

(a) $\frac{4}{9} \times 36 = 16$ girls (and 20 boys).

(b) Girls $16 + 4 = 20$ ; total $36 + 4 = 40$ . $P(\text{girl}) = \frac{20}{40} = \frac{1}{2}\text{.}$

(a) 16 girls; (b) $\frac{1}{2}$ .

Question 12

For Racial Harmony Day, the names of all 240 lower secondary pupils of a school are put in a box, and one name is drawn at random to read the pledge. The probability that the pupil drawn is in the school band is $\frac{1}{20}$ . How many of the 240 pupils are in the band?

ShowHide solution

Let there be $t$ band members. $\frac{t}{240} = \frac{1}{20}$ , so $t = \frac{240}{20} = 12$ .

Check: $\frac{12}{240} = \frac{1}{20}$ .

12 pupils.

Key Points

Favourable $= P \times$ total; total $=$ favourable $\div P$ .
With an unknown in the bag, write the probability with the unknown in the total, set it equal to the given probability, and solve.
Cross-multiply with brackets: $\frac{6}{6 + x} = \frac{2}{5}$ gives $30 = 2(6 + x)$ .
When objects are added or removed, change both the number favourable and the total.
For a spinner, angle $= P \times 360^\circ$ .
Every number of objects is a whole number from 0 to the total; a probability that does not give one is not possible.
Check every answer by putting it back into the probability.