A determinant is chosen randomly from the set of all determinant of order 4 with elements 2 and 3. The probability that determinant chosen will have an even number of 2 ’s is

Question

A determinant is chosen randomly from the set of all determinant of order 4 with elements 2 and 3. The probability that determinant chosen will have an even number of 2 ’s is (A) (215 - 2/16) (B) (215 - 2/216) (C) (28 - 1/216) (D) (215 - 1/216)

Tardigrade Admin · Accepted Answer

Order of given determinant is 4. ∴ Total number of determinants = 216 Now, number of determinants in which there are even number of 2âs = 16C2 + 16C4 + ... + 16C16 = 16C0 + 16C2 + ... + 16C16 - 16C0 = 215 - 1 ∴ Required probability = (215 - 1/216)

Q. A determinant is chosen randomly from the set of all determinant of order $4$ with elements $2$ and $3$ . The probability that determinant chosen will have an even number of $2’ s$ is

$\frac{2 ^{15} - 2}{16}$

$\frac{2 ^{15} - 2}{2 ^{16}}$

$\frac{2 ^{8} - 1}{2 ^{16}}$

$\frac{2 ^{15} - 1}{2 ^{16}}$

Q. A determinant is chosen randomly from the set of all determinant of order 4 with elements 2 and 3. The probability that determinant chosen will have an even number of 2’s is

16215−2​

216215−2​

21628−1​

216215−1​

Order of given determinant is 4. ∴ Total number of determinants =216 Now, number of determinants in which there are even number of 2’s =16C2​+16C4​+...+16C16​ =16C0​+16C2​+...+16C16​−16C0​=215−1 ∴ Required probability =216215−1​

Q. A determinant is chosen randomly from the set of all determinant of order $4$ with elements $2$ and $3$ . The probability that determinant chosen will have an even number of $2’ s$ is

$\frac{2 ^{15} - 2}{16}$

$\frac{2 ^{15} - 2}{2 ^{16}}$

$\frac{2 ^{8} - 1}{2 ^{16}}$

$\frac{2 ^{15} - 1}{2 ^{16}}$