Question 236 of 472MediumMCQ
The probability that a randomly chosen 2×2 matrix with all entries from the set of first 10 primes is singular, is equal to:
Solutions · 1
Official solution
First 10 prime numbers are
={2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
Let A is a 2 \[\times\] 2 matrix,
\[A = \left[ {(acbd) } \right]\]
Given that matrix A is singular.
\[\therefore\], A, = 0
\[ \Rightarrow \left| {(acbd) } \right| = 0\]
\[ \Rightarrow ad = bc\]
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 10C1 = 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 = 10C2 ways
And for each combination of a and d there are 4 possible values of b and c.
\[\therefore\] Total possible values = 10C2 \[\times\] 4
From case I and case II total possible values of 10 prime numbers which satisfy ad = bc condition
= 10 + 10C2 \[\times\] 4
For sample space,
Number of ways to fill element a of matrix A = chose any prime number among 10 available prime number = 10C1 ways
Similarly,
For element b of matrix A = 10C1 ways
For element c of matrix A = 10C1 ways
For element d of matrix A = 10C1 ways
\[\therefore\] Sample space = 10C1 \[\times\] 10C1 \[\times\] 10C1 \[\times\] 10C1 = 104
\[\therefore\] Probability \[ = {{10 + {{}^{10}C_{2}} \times 4} \over {{{10}^4}}}\]
\[ = {{10 + 180} \over {{{10}^4}}}\]
\[ = {{190} \over {{{10}^4}}}\]
\[ = {{19} \over {{{10}^3}}}\]
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