2014 H2 Mathematics Paper 2 Question 10 Probability
Answers (ia)
1 500 = 0.002 {\frac{1}{500}=0.002} 500 1 = 0.002 (ib)
44 125 = 0.352 {\frac{44}{125}=0.352} 125 44 = 0.352 (ic)
29 500 = 0.058 {\frac{29}{500} =0.058} 500 29 = 0.058 71 306 ≈ 0.232 {\frac{71}{306} \approx 0.232} 306 71 ≈ 0.232 Full solutions (ia)
P ( ⋆ ⋆ ⋆ ) = 1 10 × 2 10 × 1 10 = 1 500 ■ \begin{align*}
& \textrm{P}\left(\star \star \star\right) \\
&= \frac{1}{10} \times \frac{2}{10} \times \frac{1}{10} \\
&= \frac{1}{500} \; \blacksquare
\end{align*} P ( ⋆ ⋆ ⋆ ) = 10 1 × 10 2 × 10 1 = 500 1 ■ (ib)
P ( at least one ⋆ ) = 1 − P ( no ⋆ ) = 1 − 9 10 × 8 10 × 9 10 = 44 125 ■ \begin{align*}
& \textrm{P}\left(\textrm{at least one } \star\right) \\
&= 1 - \textrm{P}\left(\textrm{no } \star\right) \\
&= 1 - \frac{9}{10} \times \frac{8}{10} \times \frac{9}{10} \\
&= \frac{44}{125} \; \blacksquare
\end{align*} P ( at least one ⋆ ) = 1 − P ( no ⋆ ) = 1 − 10 9 × 10 8 × 10 9 = 125 44 ■ (ic)
P ( two × one + ) = P ( × × + ) + P ( × + × ) + P ( + × × ) ( 3 10 ) ( 1 10 ) ( 2 10 ) + ( 3 10 ) ( 3 10 ) ( 4 10 ) + ( 4 10 ) ( 1 10 ) ( 4 10 ) = 29 500 ■ \begin{align*}
& \textrm{P}\left(\textrm{two } \times \textrm{ one } +\right) \\
&= \textrm{P}\left(\times \times +\right) + \textrm{P}\left(\times + \times\right) +\textrm{P}\left(+ \times \times\right) \\
& \left(\frac{3}{10}\right)\left(\frac{1}{10}\right)\left(\frac{2}{10}\right)
+ \left(\frac{3}{10}\right)\left(\frac{3}{10}\right)\left(\frac{4}{10}\right)
+ \left(\frac{4}{10}\right)\left(\frac{1}{10}\right)\left(\frac{4}{10}\right) \\
&= \frac{29}{500} \; \blacksquare
\end{align*} P ( two × one + ) = P ( × × + ) + P ( × + × ) + P ( + × × ) ( 10 3 ) ( 10 1 ) ( 10 2 ) + ( 10 3 ) ( 10 3 ) ( 10 4 ) + ( 10 4 ) ( 10 1 ) ( 10 4 ) = 500 29 ■
(ii) P ( other two are + ∘ ∣ exactly one ⋆ ) = P ( other two are + ∘ ⋂ exactly one ⋆ ) P ( exactly one ⋆ ) \begin{align*}
&\textrm{P}\left(\textrm{ other two are } + \circ \mid \textrm{exactly one } \star\right) \\ &=\frac{\textrm{P}\left(\textrm{ other two are } + \circ \, \bigcap \, \textrm{exactly one } \star\right)}{\textrm{P}\left(\textrm{exactly one } \star\right)}
\end{align*} P ( other two are + ∘ ∣ exactly one ⋆ ) = P ( exactly one ⋆ ) P ( other two are + ∘ ⋂ exactly one ⋆ ) P ( exactly one ⋆ ) = P ( ⋆ − − ) + P ( − ⋆ − ) + P ( − − ⋆ ) = ( 1 10 ) ( 8 10 ) ( 9 10 ) + ( 9 10 ) ( 2 10 ) ( 9 10 ) + ( 9 10 ) ( 8 10 ) ( 1 10 ) = 153 500 \begin{align*}
& \textrm{P}\left(\textrm{exactly one } \star\right) \\
&= \textrm{P}\left(\star - -\right) + \textrm{P}\left(- \star -\right) + \textrm{P}\left(- - \star\right) \\
&= \left(\frac{1}{10}\right)\left(\frac{8}{10}\right)\left(\frac{9}{10}\right)
+ \left(\frac{9}{10}\right)\left(\frac{2}{10}\right)\left(\frac{9}{10}\right)
+ \left(\frac{9}{10}\right)\left(\frac{8}{10}\right)\left(\frac{1}{10}\right) \\
&= \frac{153}{500}
\end{align*} P ( exactly one ⋆ ) = P ( ⋆ − − ) + P ( − ⋆ − ) + P ( − − ⋆ ) = ( 10 1 ) ( 10 8 ) ( 10 9 ) + ( 10 9 ) ( 10 2 ) ( 10 9 ) + ( 10 9 ) ( 10 8 ) ( 10 1 ) = 500 153 P ( other two are + ∘ ⋂ exactly one ⋆ ) = + P ( ⋆ + ∘ ) + P ( + ⋆ ∘ ) + P ( ∘ ⋆ + ) = + P ( ⋆ ∘ + ) + P ( + ∘ ⋆ ) + P ( ∘ + ⋆ ) = + ( 1 10 ) ( 3 10 ) ( 3 10 ) + ( 4 10 ) ( 2 10 ) ( 3 10 ) + ( 2 10 ) ( 2 10 ) ( 2 10 ) = + ( 1 10 ) ( 4 10 ) ( 2 10 ) + ( 4 10 ) ( 4 10 ) ( 1 10 ) + ( 2 10 ) ( 3 10 ) ( 1 10 ) = 71 1000 \begin{align*}
& \textrm{P}\left(\textrm{ other two are } + \circ \, \bigcap \, \textrm{exactly one } \star\right) \\
&= \phantom{+\,} \textrm{P}\left(\star + \circ\right) + \textrm{P}\left(+ \star \circ\right) + \textrm{P}\left(\circ \star +\right) \\
& \phantom{=} + \, \textrm{P}\left(\star \circ +\right) + \textrm{P}\left(+ \circ \star\right) + \textrm{P}\left(\circ + \star\right) \\
&= \phantom{+\,} \left(\frac{1}{10}\right)\left(\frac{3}{10}\right)\left(\frac{3}{10}\right)
+ \left(\frac{4}{10}\right)\left(\frac{2}{10}\right)\left(\frac{3}{10}\right)
+ \left(\frac{2}{10}\right)\left(\frac{2}{10}\right)\left(\frac{2}{10}\right) \\
&\phantom{=} + \, \left(\frac{1}{10}\right)\left(\frac{4}{10}\right)\left(\frac{2}{10}\right)
+ \left(\frac{4}{10}\right)\left(\frac{4}{10}\right)\left(\frac{1}{10}\right)
+ \left(\frac{2}{10}\right)\left(\frac{3}{10}\right)\left(\frac{1}{10}\right) \\
&= \frac{71}{1000}
\end{align*} P ( other two are + ∘ ⋂ exactly one ⋆ ) = + P ( ⋆ + ∘ ) + P ( + ⋆ ∘ ) + P ( ∘ ⋆ + ) = + P ( ⋆ ∘ + ) + P ( + ∘ ⋆ ) + P ( ∘ + ⋆ ) = + ( 10 1 ) ( 10 3 ) ( 10 3 ) + ( 10 4 ) ( 10 2 ) ( 10 3 ) + ( 10 2 ) ( 10 2 ) ( 10 2 ) = + ( 10 1 ) ( 10 4 ) ( 10 2 ) + ( 10 4 ) ( 10 4 ) ( 10 1 ) + ( 10 2 ) ( 10 3 ) ( 10 1 ) = 1000 71 Required probability = 71 1000 ÷ 153 500 = 71 306 ■ \begin{align*}
& \textrm{Required probability} \\
&= \frac{71}{1000} \div \frac{153}{500} \\
&= \frac{71}{306} \; \blacksquare
\end{align*} Required probability = 1000 71 ÷ 500 153 = 306 71 ■