Question:
Three of the six vertices of a regular hexagon are chosen at rondom. The probability that the triangle with three vertices is equilateral, equals
Three of the six vertices of a regular hexagon are chosen at rondom. The probability that the triangle with three vertices is equilateral, equals
Updated On: Aug 1, 2022
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The Correct Option is C
Solution and Explanation
Three vertices out of 6 can be chosen in $^6C_3$ways
So, total ways =$^6C_3 =20$
Only two equilateral triangles can be formed
$\Delta$AEC and $\Delta$BFD
$\therefore \, \, \, \, \, $Favourable ways = 2
So, required probability = $\frac{2}{20}=\frac{1}{10}$
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Concepts Used:
Multiplication Theorem on Probability
In accordance with the multiplication rule of probability, the probability of happening of both the events A and B is equal to the product of the probability of B occurring and the conditional probability that event A happens given that event B occurs.
Let's assume, If A and B are dependent events, then the probability of both events occurring at the same time is given by:
\(P(A\cap B) = P(B).P(A|B)\)
Let's assume, If A and B are two independent events in an experiment, then the probability of both events occurring at the same time is given by:
\(P(A \cap B) = P(A).P(B)\)
Read More: Multiplication Theorem on Probability