JEE Main 2022 — Probability Question with Solution

First 10 prime numbers are

={2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

Let A is a 2 $$\times$$ 2 matrix,

A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]

Given that matrix A is singular.

$$\therefore$$ | A | = 0

\Rightarrow \left| {\matrix{ a & b \cr c & d \cr } } \right| = 0

$$\Rightarrow ad=bc$$

Case I :

ad = bc condition satisfy when a = b = c = d.

For ex when a = 2, b = 2, c = 2, d = 2, then ad = bc satisfy.

Now there are 10 prime numbers.

We can choose any one of the 10 prime number in $${}^{10}{C}_1$$ = 10 ways and put them in the four positions of the matrix and matrix will be singular.

$$\therefore$$ In this case, total favorable case = 10

Case 2 :

ad = bc condition satisfies when

(1)

a = 2, d - 3 then

(a) b = 2, c = 3

(b) b = 3, c = 2

or

a = 3, d = 2 then

(a) b = 2, c = 3

(b) b = 3, c = 2

So you can see for two different prime number for a and d there are 4 possible value of b and c which satisfy ad = bc condition.

Two different values of a and d can be chosen from 10 prime numbers = $${}^{10}{C}_2$$ ways

And for each combination of a and d there are 4 possible values of b and c.

$$\therefore$$ Total possible values = $${}^{10}{C}_2$$ $$\times$$ 4

From case I and case II total possible values of 10 prime numbers which satisfy ad = bc condition

= 10 + $${}^{10}{C}_2$$ $$\times$$ 4

For sample space,

Number of ways to fill element a of matrix A = chose any prime number among 10 available prime number = $${}^{10}{C}_1$$ ways

Similarly,

For element b of matrix A = $${}^{10}{C}_1$$ ways

For element c of matrix A = $${}^{10}{C}_1$$ ways

For element d of matrix A = $${}^{10}{C}_1$$ ways

$$\therefore$$ Sample space = $${}^{10}{C}_1$$ $$\times$$ $${}^{10}{C}_1$$ $$\times$$ $${}^{10}{C}_1$$ $$\times$$ $${}^{10}{C}_1$$ = 10⁴

$$\therefore$$ Probability $$=\frac{10+{}^{10}{C}_2\times 4}{{10}^4}$$

$$=\frac{10+180}{{10}^4}$$

$$=\frac{190}{{10}^4}$$

$$=\frac{19}{{10}^3}$$