A computer is programmed to play the game of 24 by independently repla

Firstly, note that both (4▲2) and (3@1) will be positive integers no matter which arithmetic operations ▲ and @ represent. Next, 24 can be broken into the product of two positive integers in the following way:

1. 1 * 24
2. 2 * 12
3. 3 * 8
4. 4 * 6
5. 6 * 4
6. 8 * 3
7. 12 * 2
8. 24 * 1

Neither (4▲2) nor (3@1) can be 1 for any of the operations, so the 1st and 8th cases are out.

(4▲2) also cannot be 3, 4, or 12. Hence, cases 3, 4, and 7 are also out.

(3@1) cannot be 12. Hence, case 2 is also out.

1. 1 * 24
2. 2 * 12
3. 3 * 8
4. 4 * 6
5. 6 * 4
6. 8 * 3
7. 12 * 2
8. 24 * 1

(4▲2) is 6 when ▲ is addition. The probability of this is given to be 0.4.
(3@1) is 4 when \(@\) is addition. The probability of this is given to be 0.6.

(4▲2) is 8 when ▲ is multiplication. The probability of this is given to be 0.2.
(3@1) is 3 when \(@\) is division OR multiplication. The probability of this is given to be 0.05 + 0.25.

Therefore, the probability that (4▲2) x (3@1) is 24 is 0.4 * 0.6 + 0.2 * (0.05 + 0.25) = 0.3.

Answer: D.­­­­­­­­