Question

A box contains $20$ cards of these $10$ have letters \[J\] printed on them and the remaining $10$ have \[E\] printed on them. $3$ Cards are drawn from the box, the probability that we can write \[JEE\] with these cards is
(A) $\dfrac{9}{80}$
(B) $\dfrac{1}{8}$
(C) $\dfrac{4}{27}$
(D) $\dfrac{15}{38}$

Answer 1

Hint: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability has been introduced in Math to predict how likely events are to happen.
Try to analyse the situation in your mind. Use notations for easily identifying the events as here there are total \[20\] cards out of which \[10\] have letter \[J\] and other remaining have \[E\] and three cards are drawn from the box, so for identifying them use the notations such as $\left( {}^{n}{{C}_{r}} \right)$ to calculate the total number of favourable outcomes. Here there are a total three numbers of picks.
Probability of an event to occur $=$$\dfrac{Possible}{Total}\dfrac{Outcomes}{Outcomes}$

Complete step-by-step solution:
Let’s first analyse the question in a better way. The given information is that the total cards in the box are $20$. Out of these cards,$10$ cards have letters \[J\] written on them and $10$ cards have letters \[E\] printed on them. Now we have to find the probability of taking out three cards in such an order that they make \[JEE\] together.
So, before we start, let’s understand what probability is. Probability is simply how likely something is to happen. And the way to express it mathematically is the number of possible outcomes upon the total number of outcomes.
$\Rightarrow $ Probability of an event to occur $=$$\dfrac{Possible}{Total}\dfrac{Outcomes}{Outcomes}$
For finding probability, first, we need to find the number of ways in which the required \[JEE\] letter can be possible. To make that pattern in three draws from the box, in the first draw the card should be picked \[J\] out of $10$ , in the second draw the card should be \[E\] out of $10$and in the last draw, the card should be \[E\] again but this time out of $9$.
We know that the total number of ways an event can occur can be calculated by combinations $\left( {}^{n}{{C}_{r}} \right)$
We have three picks,
The first pick, we want one \[J\] and ways to do that is ${}^{10}{{C}_{1}}$
The second pick, we want two \[E\] and ways to do that is ${}^{10}{{C}_{2}}$
Since we are doing these operations together, the total number of favourable outcomes will be ${}^{10}{{C}_{1}}\times $${}^{10}{{C}_{2}}$
Total number of possible choices of any three cards is ${}^{20}{{C}_{3}}$
Therefore, the required probability as per the above relation will be:
Probability $=\dfrac{{}^{10}{{C}_{1}}\times {}^{10}{{C}_{2}}}{{}^{20}{{C}_{3}}}$
Now, we can solve this to get the final probability
Probability $=\dfrac{10\times 10\times 9}{20\times 19\times 18}=\dfrac{5\times 3}{19\times 2}=\dfrac{15}{38}$

Option D is the correct answer.

Note: Try to go step by step while calculating the favourable outcomes. An alternative approach can be if you calculate the probability for all three picks separately and then multiply them together. This way you will calculate the probability for multiple events simultaneously.