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27 Feb 2016, 20:29
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
63% (02:59) correct
37%
(02:19)
wrong
based on 175
sessions
History
Date
Time
Result
Not Attempted Yet
When a certain computer plays 24-point game, each of Δ and @ representative operation of addition, subtraction, multiplication, and division. And each of the symbol has the following probability,
Δ -> +: 0.4 -:0.3×: 0.2 ÷: 0.1
@ -> +:0.6 -: 0.2 ×: 0.1 ÷: 0.1.
What is the probability if the computer calculates (4Δ2)(3@1)=24?
A. 0.28
B. 0.34
C. 0.44
D. 0.64
E. 0.84
* A solution will be posted in two days.
Δ -> +: 0.4 -:0.3×: 0.2 ÷: 0.1
@ -> +:0.6 -: 0.2 ×: 0.1 ÷: 0.1.
What is the probability if the computer calculates (4Δ2)(3@1)=24?
A. 0.28
B. 0.34
C. 0.44
D. 0.64
E. 0.84
* A solution will be posted in two days.
27 Feb 2016, 22:58
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sowragu
You are perfectly fine in two scenarios, but you are missing out on th ethird scenarioo..
1) (4*2)(3*1)=24
prob=.4*.6=.24
2)(4+2)(3+1)=24..
.2*.1=.02..
and
3) (4*2)(3/1)=24
.2*.1=.02
total= .24+.02+.02=.28..
you have missed out on third scenario..
General Discussion
27 Feb 2016, 22:52
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MathRevolution
I think Option A should be 0.26. Correct me if am wrong.
Probability of two events getting 24
For the given set {4,2} Set1 and {3,1} Set2 the maximum possible values are
{4,2} Max value = 8 (4 multiplied by 2) and Min Value = 2 (4 divided by 2)
{3,1} Max value = 4 (3 + 1) and Min Value = 2 (3 - 1)
Factors of 24 should be multiple of 8 for less than that based on the above calculation.
based on that we can arrive as below
(4*2)(3*1) = 24
(4+2)(3+1) = 24
Case1: Both SET1 AND SET2 should have the symbol as '*'. Since AND its multiplication. = .1 * .2 = 0.02
Case2: Both SET1 AND SET2 should have the symbol as '+'. Since AND its multiplication. = .6 * .4 = 0.24
Now the chance of getting 24 as outcome for the given sets are --> It should be either CASE1 or CASE2
'OR' in probability is addition. Hence = 0.24 + 0.02 = 0.26
