The probability that a randomly chosen 2 $\times$ 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to :

<p>First 10 prime numbers are</p> <p>={2, 3, 5, 7, 11, 13, 17, 19, 23, 29}</p> <p>Let A is a 2 $\times$ 2 matrix,</p> <p>$A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$</p> <p>Given that matrix A is singular.</p> <p>$\therefore$ | A | = 0</p> <p>$\Rightarrow \left| {\matrix{ a & b \cr c & d \cr } } \right| = 0$</p> <p>$\Rightarrow ad = bc$</p> <p>Case I :</p> <p>ad = bc condition satisfy when a = b = c = d.</p> <p>For ex when a = 2, b = 2, c = 2, d = 2, then ad = bc satisfy.</p> <p>Now there are 10 prime numbers.</p> <p>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.</p> <p>$\therefore$ In this case, total favorable case = 10</p> <p>Case 2 :</p> <p>ad = bc condition satisfies when</p> <p>(1)</p> <p>a = 2, d - 3 then</p> <p>(a) b = 2, c = 3</p> <p>(b) b = 3, c = 2</p> <p>or</p> <p>a = 3, d = 2 then</p> <p>(a) b = 2, c = 3</p> <p>(b) b = 3, c = 2</p> <p>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.</p> <p>Two different values of a and d can be chosen from 10 prime numbers = ${}^{10}C_{2}$ ways</p> <p>And for each combination of a and d there are 4 possible values of b and c.</p> <p>$\therefore$ Total possible values = ${}^{10}C_{2}$ $\times$ 4</p> <p>From case I and case II total possible values of 10 prime numbers which satisfy ad = bc condition</p> <p>= 10 + ${}^{10}C_{2}$ $\times$ 4</p> <p>For sample space,</p> <p>Number of ways to fill element a of matrix A = chose any prime number among 10 available prime number = ${}^{10}C_{1}$ ways</p> <p>Similarly,</p> <p>For element b of matrix A = ${}^{10}C_{1}$ ways</p> <p>For element c of matrix A = ${}^{10}C_{1}$ ways</p> <p>For element d of matrix A = ${}^{10}C_{1}$ ways</p> <p>$\therefore$ Sample space = ${}^{10}C_{1}$ $\times$ ${}^{10}C_{1}$ $\times$ ${}^{10}C_{1}$ $\times$ ${}^{10}C_{1}$ = 10⁴</p> <p>$\therefore$ Probability $= {{10 + {{}^{10}C_{2}} \times 4} \over {{{10}^4}}}$</p> <p>$= {{10 + 180} \over {{{10}^4}}}$</p> <p>$= {{190} \over {{{10}^4}}}$</p> <p>$= {{19} \over {{{10}^3}}}$</p>