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Find the chance

(1) that two specified persons obtain opposite seats,

(2) that they obtain adjacent seats on the same side.

Answer

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Hint: Whenever a question is given of this sort where we are supposed to find the chances of an event happening, we use probability.

Let us consider 2 people X and Y.

The condition is the seats should be opposite to each other and there will be only one seat opposite to X.

Now, the probability that Y sits opposite to \[X = \dfrac{6}{7} \times \dfrac{1}{7} = \dfrac{1}{7}\]

So the answer for the first part \[ = \dfrac{1}{7}\]

2) Total there are 7 persons, we have to select 2 adjacent seats that can be done using combination.

Therefore,

The seats can be selected in ${}^7{C_2}ways = 21ways$

Note: ${}^7{C_2}$ is the total ways in which the seat can be taken by the two people, it is not the final answer. The final answer will be the probability of the favourable condition divided by the total outcomes.

Now, X and Y can be seated adjacent in 4 ways, which is our favourable condition.

Now, to find the probability we are going to use the formula of probability

Therefore,

\[Required{\text{ }}Probability = \dfrac{{Favourable\,Condition}}{{Total\,Conditions}} = \dfrac{4}{{21}}\]

Let us consider 2 people X and Y.

- 1) Now, the probability of X getting a seat \[ = \dfrac{6}{7}\]

The condition is the seats should be opposite to each other and there will be only one seat opposite to X.

Now, the probability that Y sits opposite to \[X = \dfrac{6}{7} \times \dfrac{1}{7} = \dfrac{1}{7}\]

So the answer for the first part \[ = \dfrac{1}{7}\]

Therefore,

The seats can be selected in ${}^7{C_2}ways = 21ways$

Note: ${}^7{C_2}$ is the total ways in which the seat can be taken by the two people, it is not the final answer. The final answer will be the probability of the favourable condition divided by the total outcomes.

Now, X and Y can be seated adjacent in 4 ways, which is our favourable condition.

Now, to find the probability we are going to use the formula of probability

Therefore,

\[Required{\text{ }}Probability = \dfrac{{Favourable\,Condition}}{{Total\,Conditions}} = \dfrac{4}{{21}}\]