# Probability : Problems using Tree Diagrams and Tables

1.     A family has $\displaystyle 4$ children. Draw a tree diagram to list all possible outcomes. If each outcomes is equally likely to occur, find the probability that the last two children are girls. Find also the probability that exactly two children are boys.

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Let Boy be denoted by $\displaystyle B$ and Girl be by $\displaystyle G.$

$\displaystyle \therefore\ \$Number of possible outcomes $\displaystyle =16$

Set of favourable outcomes for the last two children are girls

$\displaystyle \ \ \ \ \ \ \ \ = \{(B,B,G,G),(B,G,G,G),(G,B,G,G),(G,G,G,G)\}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle =4$

$\displaystyle \therefore\ \ P$( the last two children are girls) $\displaystyle = \frac{4}{16}= \frac{1}{4}$

Set of favourable outcomes for exactly two children are boys

$\displaystyle \ \ \ \ \ \ \ \ = \{(B,B,G,G),(B,G,B,G),(B,G,G,B),(G,B,B,G),(G,B,G,B),(G,G,B,B)\}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle =6$

$\displaystyle \therefore\ \ P$( the last two children are girls) $\displaystyle = \frac{6}{16}= \frac{3}{8}$

2.      How many $\displaystyle 2$-digit numerals can you form from $\displaystyle 2,3,5,6$ without repeating any digit? Find the probability of a numeral which is divisible by $\displaystyle 2$. If one of these numerals is chosen at random, find the probability that it is divisible by $\displaystyle 13.$

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$\displaystyle \therefore\ \$Number of possible outcomes $\displaystyle =12$

Set of favourable outcomes for a numeral which is divisible by $\displaystyle 2$

$\displaystyle \ \ \ \ \ \ \ \ = \{26,32,36,52,56,62\}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle =6$

$\displaystyle \therefore\ \ P$( a numeral which is divisible by $\displaystyle 2$) $\displaystyle = \frac{6}{12}= \frac{1}{2}$

Set of favourable outcomes for a numeral which is divisible by $\displaystyle 13$

$\displaystyle \ \ \ \ \ \ \ \ = \{26,52,65\}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle =3$

$\displaystyle \therefore\ \ P$( a numeral which is divisible by $\displaystyle 13$) $\displaystyle = \frac{3}{12}= \frac{1}{4}$

3.      How many $\displaystyle 3$-digit numerals can you form from $\displaystyle 1, 0 , 5$ and $\displaystyle 6$ without repeating any digit? Find the probability of numeral which is divisible by $\displaystyle 5.$

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$\displaystyle \therefore\ \$Number of possible outcomes $\displaystyle =18$

Set of favourable outcomes for a numeral which is divisible by $\displaystyle 5$

$\displaystyle \ \ \ \ \ \ \ \ = \{105,150,160,165,510,560,605,610,615,650\}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle =10$

$\displaystyle \therefore\ \ P$( a numeral which is divisible by $\displaystyle 5$) $\displaystyle = \frac{10}{18}= \frac{5}{9}$

4.      A fair coin is tossed three times. Draw a tree diagram to determine the set of all possible outcomes. Hence find the probability of

(b) getting two heads and one tail in any order.

(c) getting at least one head.

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$\displaystyle \therefore\ \$ Set of possible outcomes $\displaystyle = \{HHH,HHT,HTH, HTT,THH,THT,TTH,TTT\}$

$\displaystyle \therefore\ \$Number of possible outcomes $\displaystyle = 8$

(a)       Set of favourable outcomes for getting three heads $\displaystyle = \{HHH \}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle = 1$

$\displaystyle \therefore\ \ P$(getting three heads) $\displaystyle = \frac{1}{8}$

(b)       Set of favourable outcomes for getting two heads and one tail in any order
$\displaystyle \ \ \ \ \ \ = \{HHT,HTH,THH \}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle = 3$

$\displaystyle \therefore\ \ P$(getting two heads and one tail in any order) $\displaystyle = \frac{3}{8}$

(c)       Set of favourable outcomes for getting at least one head
$\displaystyle \ \ \ \ \ \ = \{HHH,HHT,HTH, HTT,THH,THT,TTH \}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle = 7$

$\displaystyle \therefore\ \ P$(getting at least one head) $\displaystyle = \frac{7}{8}$

(d)    Set of favourable outcomes for getting no head
$\displaystyle \ \ \ \ \ \ = \{TTT \}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle = 1$

$\displaystyle \therefore\ \ P$(getting no head) $\displaystyle = \frac{1}{8}$

5.      Draw a table of all posssible outcomes for throwing two fair dice. Hence, calculate the probability that

(a) the product of the scores on the two dice is divisible by $\displaystyle 6$ or $\displaystyle 9.$

(b) both dice showing the same number.

(c) the sum of the score is a multiple of $\displaystyle 3.$

(d) the total score on the two dice is prime.

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$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ \ \ \ {{\text{2}}^{{\text{nd}}}}\ \text{die}\\ \ {{\text{1}}^{{\text{st}}}}\ \text{die}\ \ \ \begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & {(1,1)} & {(1,2)} & {(1,3)} & {(1,4)} & {(1,5)} & {(1,6)} \\ \hline 2 & {(2,1)} & {(2,2)} & {(2,3)} & {(2,4)} & {(2,5)} & {(2,6)} \\ \hline 3 & {(3,1)} & {(3,2)} & {(3,3)} & {(3,4)} & {(3,5)} & {(3,6)} \\ \hline 4 & {(4,1)} & {(4,2)} & {(4,3)} & {(4,4)} & {(4,5)} & {(4,6)} \\ \hline 5 & {(5,1)} & {(5,2)} & {(5,3)} & {(5,4)} & {(5,5)} & {(5,6)} \\ \hline 6 & {(6,1)} & {(6,2)} & {(6,3)} & {(6,4)} & {(6,5)} & {(6,6)} \\ \hline\end{array}\end{array}$

$\displaystyle \therefore\ \$Number of possible outcomes $\displaystyle =36$

(a)     Set of favourable outcomes for the product of the scores on the two dice is divisible by $\displaystyle 6$ or $\displaystyle 9$

$\displaystyle \ \ \ \ \ \ \ \ = \{(1,6),(2,3), (2,6), (3,2), (3,3), (3, 4), (3, 6),(4,3), \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4, 6), (5,6), (6,1), (6,2), (6, 3), (6, 4), (6, 5), (6, 6)\}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle =16$

$\displaystyle \therefore\ \ P$(the product of the scores on the two dice is divisible by $\displaystyle 6$ or $\displaystyle 9$) $\displaystyle = \frac{16}{36}= \frac{4}{9}$

(b)     Set of favourable outcomes for both dice showing the same number

$\displaystyle \ \ \ \ \ \ \ \ = \{(1,1),(2,2), (3,3), (4,4), (5,5), (6,6) \}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle = 6$

$\displaystyle \therefore\ \ P$(both dice showing the same number) $\displaystyle = \frac{6}{36}= \frac{1}{6}$

(c)     Set of favourable outcomes for the sum of the score is a multiple of $\displaystyle 3$

$\displaystyle \ \ \ \ \ \ \ \ = \{(1,2),(1,5), (2,1), (2,4), (3,3),(3,6),\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4,2), (4,5), (5,1), (5,4), (6,3), (6,6) \}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle = 12$

$\displaystyle \therefore\ \ P$(both dice showing the same number) $\displaystyle = \frac{12}{36}= \frac{1}{3}$

(d)     Set of favourable outcomes for the total score on the two dice is prime

$\displaystyle \ \ \ \ \ \ \ \ = \{(1,1),(1,2), (1,4),(1,6),(2,1), (2,3), (2,5), \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5) \}$

$\displaystyle \therefore\ \$Number of favourable outcomes $\displaystyle = 15$

$\displaystyle \therefore\ \ P$(both dice showing the same number) $\displaystyle = \frac{15}{36}= \frac{5}{12}$