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

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

& \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{aligned}$$ Let us assume that $${{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{aligned} & \Rightarrow P\left( {{L}_{1}} \right)=1-\dfrac{3}{5} \\\ & \Rightarrow P\left( {{L}_{1}} \right)=\dfrac{2}{5} \\\ \end{aligned}$$ We are given that the chances of team winning on Sunday as 35%. Let us assume that $${{W}_{2}}$$ be the event of winning on Sunday then the probability is given as $$\begin{aligned} & \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{aligned}$$ Let us assume that $${{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{aligned} & \Rightarrow P\left( {{L}_{2}} \right)=1-\dfrac{7}{20} \\\ & \Rightarrow P\left( {{L}_{2}} \right)=\dfrac{13}{20} \\\ \end{aligned}$$ Let us assume that ‘E’ be the event of losing both the matches. We know that, if the probability of occurring the events A, B are$$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{aligned} & \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{aligned}$$ Therefore, the probability of losing both the matches as is 26%. **So, the correct answer is “Option d”.** **Note:** Students may make mistakes in finding the required probability. We know that, if the probability of occurring the events A, B are$$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.