When a certain computer plays 24-point game each of # and @ representative operation of addition, subtraction, multiplication and division. And each of the symbol has the following probability,
# -> +: 0.4 -:0.3×: 0.2 ÷: 0.1
@ -> +:0.6 -: 0.2 ×: 0.1 ÷: 0.1
What is the probability if the computer calculate (4#2)(3@1)=24?
A. 0.28
B. 0.34
C. 0.44
D. 0.64
E. 0.84
The OA is A.
I'm really confused with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.
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When a certain computer plays 24-point game...
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BTGmoderatorLU
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This is quite an onerous task.
Listing outcomes gives:
(4+2)(3+1) = 24
(4+2)(3x1) = 18
(4+2)(3/1) = 18
(4+2)(3-1) = 12
(4-2)(3+1) = 8
(4-2)(3x1) = 6
(4-2)(3/1) = 6
(4-2)(3-1) = 4
(4x2)(3+1) = 32
(4x2)(3x1) = 24
(4x2)(3/1) = 24
(4x2)(3-1) = 16
(4/2)(3+1) = 8
(4/2)(3x1) = 6
(4/2)(3/1) = 6
(4/2)(3-1) = 4
There are exactly 3 successful outcomes of 24, with probabilities P, Q and R:
(4+2)(3+1) = 24, P = 0.4 x 0.6 = 0.24
(4x2)(3x1) = 24, Q = 0.2 x 0.1 = 0.02
(4x2)(3/1) = 24, R = 0.2 x 0.1 = 0.02
Probability (P or Q or R) = P + Q + R = 0.28
Listing outcomes gives:
(4+2)(3+1) = 24
(4+2)(3x1) = 18
(4+2)(3/1) = 18
(4+2)(3-1) = 12
(4-2)(3+1) = 8
(4-2)(3x1) = 6
(4-2)(3/1) = 6
(4-2)(3-1) = 4
(4x2)(3+1) = 32
(4x2)(3x1) = 24
(4x2)(3/1) = 24
(4x2)(3-1) = 16
(4/2)(3+1) = 8
(4/2)(3x1) = 6
(4/2)(3/1) = 6
(4/2)(3-1) = 4
There are exactly 3 successful outcomes of 24, with probabilities P, Q and R:
(4+2)(3+1) = 24, P = 0.4 x 0.6 = 0.24
(4x2)(3x1) = 24, Q = 0.2 x 0.1 = 0.02
(4x2)(3/1) = 24, R = 0.2 x 0.1 = 0.02
Probability (P or Q or R) = P + Q + R = 0.28
