Question

If the lengths of sides of a triangle are decided by the three throws of single fair die, then the probability that the triangle is of maximum area gives that is an isosceles triangle, is :

• A. $\frac{1}{26}$
• B. $\frac{1}{27}$
• C. $\frac{1}{21}$
• D. $\frac{1}{15}$

Answer 1

The correct option is C $\frac{1}{21}$

Sol :- Given,

when lengths of sides of a triangle are decided

by the three throws of single fair die.

then, asked to find out the probability that the

triangle is of maximum area gives an isosceles $△$,

will be __

so, as already mentioned in the data, do not

consider an equilateral $△$ , rather consider an isosceles

$△$.

And for an isosceles $△$ there occurs 21 cases and

each case will be occurring thrice i.e., as follows,

$(2,2,1)(2,2,3)(3,3,1)(3,3,2)(3,3,4)(3,3,5)$

$(4,4,1)(4,4,2)(4,4,3)(4,4,5)(4,4,6)(5,5,1)$

$(5,5,2)(5,5,3)(5,5,4)(5,5,6)(6,6,1)(6,6,2)$

$(6,6,3)(6,6,4)(6,6,5)$ and these are all the

probabilities of an isosceles triangle for the three throws

of die.

From the above probabilities (6,6,5) has the maximum

area.

$P\left(A\right)=\frac{n\left(E\right)}{n\left(S\right)}$

n(E) = no. of cases.

$\Rightarrow P\left(A\right)=\frac{3}{63}=\frac{1}{21}$

n(S) = Total no. of cases.

$\therefore$ the probability that the $△$ is of maximum area

for an isosceles triangle will be $\frac{1}{21}$

Answer is option [C]