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Question: A basket contains 8 Red and 6 Pink toys. There is another basket which contains 7 Red and 8 Pink toy...

A basket contains 8 Red and 6 Pink toys. There is another basket which contains 7 Red and 8 Pink toys. One toy is to be drawn from either of the two baskets. What is the probability of drawing a Pink toys?

Explanation

Solution

We will start by using the concepts of Probability. We will use the multiplication theorem of probability to find the probability of Pink toys. For this we will first find the probability of choosing a basket and then we will find the probability of drawing a pink toy from it and multiply both. Doing similarly for both the baskets we will find the probability of drawing a pink toy by adding them.

Complete step-by-step answer:
Now, we have been given that a basket contains 8 Red and 6 Pink toys. There is another basket which contains 7 Red and 8 Pink toys.
Now, we know that the probability of an event A is,
P(A)=Number of favourable outcomesTotal possible outcomesP\left( A \right)=\dfrac{Number\ of\ favourable\ outcomes}{Total\ possible\ outcomes}
Now, we have the probability of picking a toy from the first basket =12=P(A)=\dfrac{1}{2}=P\left( A \right).
Now, we have the probability of picking pink toy from first basket,
=Number of pink toysTotal number of toys×P(A) =614×12 =314...........(1) \begin{aligned} & =\dfrac{Number\ of\ pink\ toys}{Total\ number\ of\ toys}\times P\left( A \right) \\\ & =\dfrac{6}{14}\times \dfrac{1}{2} \\\ & =\dfrac{3}{14}...........\left( 1 \right) \\\ \end{aligned}
We have multiplied by P(A) because by multiplication theorem if E and F are two events associated with a sample space S then the probability of simultaneously occurrence is,
P(EF)=P(E)×P(FE)P\left( E\cap F \right)=P\left( E \right)\times P\left( \dfrac{F}{E} \right)
Similarly, we have the probability of picking a toy from the second basket =P(B)=12=P\left( B \right)=\dfrac{1}{2}.
So, we have the probability of picking pink toy from second basket,
=815×12 =415...........(2) \begin{aligned} & =\dfrac{8}{15}\times \dfrac{1}{2} \\\ & =\dfrac{4}{15}...........\left( 2 \right) \\\ \end{aligned}
Now, we have from (1) and (2) the probability of drawing a pink toy,
=314+415 =3×15+4×1414×15 =101210 \begin{aligned} & =\dfrac{3}{14}+\dfrac{4}{15} \\\ & =\dfrac{3\times 15+4\times 14}{14\times 15} \\\ & =\dfrac{101}{210} \\\ \end{aligned}

Note: It is important to note that we have used the multiplication theorem to find the probability of drawing a pink toy from each basket wherein we have first found the probability of selecting a basket. Then we have multiplied it by the probability of picking a pink toy from that basket. Another silly mistake that the students can make is by assuming the number of red toys to be 6 and pink toys to be 8 and this might result in the wrong answer. So, they must read the question carefully, write down the given data and then apply them in the formula.