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.­­­­­­­­

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Because (4▲2) x (3@1) is a product of two factors, first of all, look for the product of two factors that give us 24.

1*24, 2*12, 3*8 and 4*6

Neither term can be made made more than 8 (= 4*2) hence ignore the first two products.

(4 * 2) x (3 * 1) = 8 * 3
(4 * 2) x (3 ÷ 1) = 8 * 3
(4 + 2) x (3 + 1) = 6 * 4

Probability of converting ▲ to x AND of converting to x or ÷ will be = 0.2 * 0.3 = 0.06
Probability of converting ▲ to + AND of converting to + will be = 0.4 * 0.6 = 0.24

Total probability = 0.06 + 0.24 = 0.3

Answer (D)

 

One of the best problems I've seen in a while. Just list out the possibilities:

Given: (4_2) * (3_1) = 24?

(4 x 2) * (3 x 1) = 8 x 3 = 24
(4 x 2) * ( 3 ÷ 1) = 8 x 3 = 24
(4 + 2) * (3 + 1) = 6 x 4 = 24

For 1 = 0.2 x 0.25 = 0.05
For 2 = 0.2 x 0.05 = 0.01
For 3 = 0.6 x 0.4 = 0.24

Total = 1 or 2 or 3 ways = 0.05 + 0.01 + 0.24 = 0.06 + 0.24 = 0.3

big thing with this is staying organized and structured.

Quite a good one actually. To prepare for such questions, I share one method with all my students stating that even though you know the formula you must practice writing factors by making pairs of numbers whose product results in the number whose factor we are finding
e.g.

24 =
1*24
2*12
3*8
4*6

Now, we need to think which case could be equated with (4▲2) x (3@1)

(4▲2) as per our result could either be 4x2 = 8 or 4+2=6 or 4÷2 = 2 or 4-2 = 2 so, we have 4 cases here

Case 1: (4▲2) = 4x2 = 8 the probability of this to happen is 0.2
in which case (3@1) must be 3 i.e. 3x1 or 3÷1 = 3 [because 8*3 = 24]
Probability of (3@1) to become 3x1 or 3÷1 = 0.25+0.05 = 0.30
So total Probability = 0.2*0.3 = 0.06

Case 2: (4▲2) = 4+2 = 6 the probability of this to happen is 0.4
in which case (3@1) must be 4 i.e. 3+1 = 4 [because 6*4 = 24]
Probability of (3@1) to become 3+1 = 0.6
So total Probability = 0.4*0.6 = 0.24

Case 3: (4▲2) = 4[color=#e86e04]÷2 = 2 the probability of this to happen is [/color]0.1
in which case (3@1) must be 12 BUT this is NOT POSSIBLE

Case 4: (4▲2) = 4-2 = 2 the probability of this to happen is 0.3
in which case (3@1) must be 12 BUT this is NOT POSSIBLE


So Final Probrbaility = Case 1 + Case 2 = 0.06+0.24 = 0.3
ANswer: Option D

Related Video:

The given expression: (4▲2) x (3@1)

Observe that whatever mathematical operation replaces ▲ and @, (4▲2) and (3@1) will both be integers.

So, when can the product of 2 integers get us 24?

(4▲2) x (3@1) = 24

4 possible number pairs -> (1,24) (2,12) (3,8) (4,6)

Look at (4▲2) and (3@1). No mathematical operation on any of these terms can give us "24" or even "12". Even 4 x 2 gives us a max of 8.

So, we can reject (1,24) (2,12).

We are left with 2 possible number pairs -> (3,8) (4,6)

First, let's focus on (3,8).

No operation on (3@1) can yield 8. But 4 x 2 can give us 8.

This leads to two cases ->

(1) (4 x 2) x (3 x 1) -> 8 x 3
(2) (4 x 2) x (3 ÷ 1) -> 8 x 3


If we look at (4,6) ->

No operation on (3@1) can yield 6. But 4+2 can give us 6.

Here, there is one case ->

(3) (4+2) x (3+1) -> 6 x 4

Any of these 3 cases are possible. Thus, the probability that the computer wins is the probability of any one of these 3 cases happening.

P(computer wins) = P(24 achieved) = P(case 1) OR P(case 2) OR P(case 3)

= (0.2 x 0.25) + (0.2 x 0.05) + (0.4 x 0.6)
= 0.05 + 0.01 + 0.24
= 0.06 + 0.24
= 0.30

Choice D.

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Harsha