Question

A professional baseball team will play 1 game Saturday and 1 game on Sunday. A sportswriter estimates that the team has 60% chance of winning on Saturday and 35% chance of winning on Sunday. Using the sportswriter estimation, what is the probability that the team will lose both the matches?
(a) 12%
(b) 21%
(c) 25%
(d) 26%

Answer 1

Hint: We solve this problem first by finding the probability of losing on Saturday and Sunday. If
\[P\left( E \right)\] be the probability of occurring the event ‘E’ then the probability of not occurring the event is given as
\[\Rightarrow P\left( {{E}'} \right)=1-P\left( E \right)\]
After finding the probability of losing the match on Saturday and Sunday we use permutations to find the probability of losing both matches. If probability of occurring the events A, B are
\[P\left( A \right),P\left( B \right)\] then probability of occurring both events is given as
\[\Rightarrow P\left( E \right)=P\left( A \right)\times P\left( B \right)\]

Complete step by step answer:
We are given that the chances of the team winning on Saturday was 60%.
Let us assume that \[{{W}_{1}}\] be the event of winning on Saturday then the probability is given as
\[\begin{align}
  & \Rightarrow P\left( {{W}_{1}} \right)=60\% \\
 & \Rightarrow P\left( {{W}_{1}} \right)=\dfrac{60}{100} \\
 & \Rightarrow P\left( {{W}_{1}} \right)=\dfrac{3}{5} \\
\end{align}\]
Let us assume that \[{{L}_{1}}\] be the event of losing the match on Saturday.
We know that if \[P\left( E \right)\] be the probability of occurring the event ‘E’ then the probability of not occurring the event is given as
\[\Rightarrow P\left( {{E}'} \right)=1-P\left( E \right)\]
By using the above formula we get the probability of losing the match on Saturday as
\[\Rightarrow P\left( {{L}_{1}} \right)=1-P\left( {{W}_{1}} \right)\]
By substituting the required values we get
\[\begin{align}
  & \Rightarrow P\left( {{L}_{1}} \right)=1-\dfrac{3}{5} \\
 & \Rightarrow P\left( {{L}_{1}} \right)=\dfrac{2}{5} \\
\end{align}\]
We are given that the chances of team winning on Sunday as 35%.
Let us assume that \[{{W}_{2}}\] be the event of winning on Sunday then the probability is given as
\[\begin{align}
  & \Rightarrow P\left( {{W}_{2}} \right)=35\% \\
 & \Rightarrow P\left( {{W}_{2}} \right)=\dfrac{35}{100} \\
 & \Rightarrow P\left( {{W}_{2}} \right)=\dfrac{7}{20} \\
\end{align}\]
Let us assume that \[{{L}_{2}}\] be the event of losing the match on Sunday.
We know that if \[P\left( E \right)\] be the probability of occurring the event ‘E’ then the probability of not occurring the event is given as
\[\Rightarrow P\left( {{E}'} \right)=1-P\left( E \right)\]
By using the above formula we get the probability of losing the match on Sunday as
\[\Rightarrow P\left( {{L}_{2}} \right)=1-P\left( {{W}_{2}} \right)\]
By substituting the required values we get
\[\begin{align}
  & \Rightarrow P\left( {{L}_{2}} \right)=1-\dfrac{7}{20} \\
 & \Rightarrow P\left( {{L}_{2}} \right)=\dfrac{13}{20} \\
\end{align}\]
Let us assume that ‘E’ be the event of losing both the matches.
We know that, if the probability of occurring the events A, B are\[P\left( A \right),P\left( B \right)\] then probability of occurring both events is given as
\[\Rightarrow P\left( E \right)=P\left( A \right)\times P\left( B \right)\]
By using the above formula we get the probability of losing both the matches as
\[\begin{align}
  & \Rightarrow P\left( E \right)=P\left( {{L}_{1}} \right)\times P\left( {{L}_{2}} \right) \\
 & \Rightarrow P\left( E \right)=\dfrac{2}{5}\times \dfrac{13}{20} \\
 & \Rightarrow P\left( E \right)=\dfrac{26}{100}=26\% \\
\end{align}\]
Therefore, the probability of losing both the matches as is 26%.

So, the correct answer is “Option d”.

Note: Students may make mistakes in finding the required probability.
We know that, if the probability of occurring the events A, B are\[P\left( A \right),P\left( B \right)\] then probability of occurring both events is given as
\[\Rightarrow P\left( E \right)=P\left( A \right)\times P\left( B \right)\]
This formula is obtained from the permutations.
But students may do mistake and use the combinations instead of permutations and take the formula as
\[\Rightarrow P\left( E \right)=P\left( A \right)+P\left( B \right)\]
This gives the wrong answer.