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# There are 2 red flowers and 2 yellow flowers in a basket. A child picks up at random three flowers. What is the probability of picking up both the yellow flowers?

Last updated date: 19th Sep 2024
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Hint: Here, we need to find the probability of picking up both the yellow flowers. First, we will use combinations to find the total number of outcomes and the number of favourable outcomes. Then, we will use the formula for probability of an event to find the probability of picking up both the yellow flowers.

Formula Used:
We will use the following formulas:
1.The number of ways in which the $n$ objects can be kept in $r$ spaces, where order of objects is not important or does not matter, can be found using the formula for combinations ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
2.The probability of an event is given by $P\left( E \right) = \dfrac{{{\rm{Number\, of\, favourable\, outcomes}}}}{{{\rm{Number\, of\, total\, outcomes}}}}$.

We will use combinations to find the number of favourable outcomes and unfavourable outcomes.
First, we will find the number of total outcomes.
The child picks three flowers from the four flowers.
Therefore, the number of ways in which the child can pick 3 flowers from 4 flowers is given by ${}^4{C_3}$.
Substituting $n = 4$ and $r = 3$ in the formula for combinations ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$\Rightarrow {}^4{C_3} = \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}}$
Subtracting the terms in the expression, we get
$\Rightarrow {}^4{C_3} = \dfrac{{4!}}{{3!1!}}$
Simplifying the expression, we get
$\Rightarrow {}^4{C_3} = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 1}}$
Therefore, we get
$\Rightarrow {}^4{C_3} = 4$
Therefore, the number of ways in which the child can pick 3 flowers from 4 flowers is 4 ways.
Thus, the total number of possible outcomes is 4.
Now, we will find the number of favourable outcomes.
The child picks three flowers from the four flowers.
The favourable event is that the child picks both the yellow flowers.
This means that out of the 3 flowers picked, 2 flowers are yellow, and 1 flower is red.
The number of ways in which the child can pick 1 red flower from 2 red flowers is given by ${}^2{C_1}$.
The number of ways in which the child can pick 2 yellow flowers from 2 yellow flowers is given by ${}^2{C_2}$.
Therefore, the number of ways in which the child can pick 1 red flower and 2 yellow flowers from the 4 flowers is given by ${}^2{C_1} \times {}^2{C_2}$.
Using the formula for combinations ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$\Rightarrow {}^2{C_1} \times {}^2{C_2} = \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}} \times \dfrac{{2!}}{{2!\left( {2 - 2} \right)!}}$
Subtracting the terms in the expression, we get
$\Rightarrow {}^2{C_1} \times {}^2{C_2} = \dfrac{{2!}}{{1!1!}} \times \dfrac{{2!}}{{2!0!}}$
The value of $0!$ is 1.
Simplifying the expression, we get
$\Rightarrow {}^2{C_1} \times {}^2{C_2} = \dfrac{{2 \times 1}}{{1 \times 1}} \times \dfrac{{2 \times 1}}{{2 \times 1 \times 1}}$
Therefore, we get
$\begin{array}{l} \Rightarrow {}^2{C_1} \times {}^2{C_2} = 2 \times 1\\ \Rightarrow {}^2{C_1} \times {}^2{C_2} = 2\end{array}$
Therefore, the number of ways in which the child can pick 1 red flower and 2 yellow flowers from the 4 flowers is 2 ways.
Thus, the number of favourable outcomes is 2.
Finally, we will find the probability of picking both the yellow flowers when 3 flowers are picked.
The probability of an event is given by $P\left( E \right) = \dfrac{{{\rm{Number\, of\, favourable\, outcomes}}}}{{{\rm{Number\, of\, total\, outcomes}}}}$.
Substituting the number of favourable outcomes as 2 and the number of total outcomes as 4, we get the required probability as
Probability of picking both the yellow flowers $= \dfrac{2}{4} = \dfrac{1}{2}$
Therefore, the probability of picking both the yellow flowers is $\dfrac{1}{2}$.

Note: We used combinations instead of permutations to find the number of ways to pick the flowers from the basket. This is because the order does not matter. Suppose that the red flowers are A and B. If he picks A first, and then B, he picks two red flowers. If he picks B first, then A, he picks two red flowers. If we used permutations, the formula would count the above two events as 2 separate ways of picking the two red flowers. Since we are only concerned with the type of flowers he picks, we do not need the order in which he picks them. Hence, we used combinations.