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5 out of 6 people who usually work in an office prefer coffee in the mid morning, the others always drink tea. This morning of the usual 6, only 3 are present. The probability that one of them drinks tea is?
$
  A.\,\dfrac{1}{2} \\
  B.\;\dfrac{1}{{12}} \\
  C.\;\dfrac{5}{{72}} \\
  D.\;\dfrac{5}{{72}} \\
 $

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Last updated date: 20th Jun 2024
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Answer
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Hint:
This is a standard type of question based on fundamental theory of probability. First compute the total number of ways for selecting 3 out of 6 people. Then compute the favourable ways, which will include 1 person as sure for tea and other 2 out of 5 persons for coffee. Finally apply a fundamental formula to find out the probability.

Complete step by step solution:
There are a total of 6 people in the office and 5 of them prefer coffee and 1 always drink tea. Here we will use combinatorial formula to find the number of ways for selection of person as coffee lover and tea lover.
Now, on that day only 3 were present.
So, total number of ways to have 3 person out of 6 will be,
Total ways of presence = $^6{C_3}$
Now favourable number of ways to have 3 person will be,
1 person as sure tea lover $ \times $ 2 more person from remaining 5 persons
\[{ = ^1}{C_1}{ \times ^5}{C_2}\], need to multiply as both will be required together.
Thus we have number of favourable ways \[{ = ^1}{C_1}{ \times ^5}{C_2}\]
Since,
Probability = \[\dfrac{{favourable\;number\;of\;ways}}{{total\;number\;of\;ways}}\]
Now substitute the values in above, we get
Probability =
$
= \dfrac{{^1{C_1}{ \times ^5}{C_2}}}{{^6{C_3}}} \\
= \dfrac{{1 \times \dfrac{{5!}}{{2! \times 3!}}}}{{\dfrac{{6!}}{{3! \times 3!}}}} \\
= \dfrac{{10}}{{20}} \\
= \dfrac{1}{2} \\
$ ( $^1{C_1}$ will be 1)

$\therefore $ The probability that one of them drinks tea is $\dfrac{1}{2}$.

Note:
Probability refers to the possibility of occurrence of some event. It is a branch of mathematics that dealing with possibilities and representing is numerical. The probability is equal to the ratio of the number of favourable outcomes and the total number of outcomes. Thus,
Probability of event to happen = Number of favourable outcomes / Total Number of outcomes possible.