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18 Nov 2019, 07:50
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
61% (01:34) correct
39%
(01:54)
wrong
based on 244
sessions
History
Date
Time
Result
Not Attempted Yet
Box A and B each contain five cards numbered 1, 2, 3, 4 and 5. If one card is to be taken out from box A and B, what is the probability the product of the numbers in two cards will be less than 4?
(A) 2/25
(B) 4/25
(C) 1/5
(D) 1/4
(E) 2/5
(A) 2/25
(B) 4/25
(C) 1/5
(D) 1/4
(E) 2/5
18 Nov 2019, 12:13
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To find the answer, find the number of combinations that will result in a product less than 4. in this case, opposites will count as separate draws, ie, 1*2 and 2*1 are 2 separate combinations.
1*1, 1*2, 1*3,2*1, and 3*1
there are 5 different combinations that qualify. to find probability, divide by the total number of combinations possible (5*5)=25
5/25=1/5
Ans=C
1*1, 1*2, 1*3,2*1, and 3*1
there are 5 different combinations that qualify. to find probability, divide by the total number of combinations possible (5*5)=25
5/25=1/5
Ans=C
20 Nov 2019, 08:52
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Refer Attachment
Set A has 1 2 3 4 5
Set B has 1 2 3 4 5
So total 5x5=25 combinations
Now for product of any 2 to be less than 4
Set A becomes 1 2 34 5
Similarly Set B becomes 1 2 34 5
Combinations possible from the remaining
1x1 1x2 1x3 2x1 and 3x1
Other combinations does not meet the requirements
So the answer is 5/25 = 1/5
Set A has 1 2 3 4 5
Set B has 1 2 3 4 5
So total 5x5=25 combinations
Now for product of any 2 to be less than 4
Set A becomes 1 2 3
Similarly Set B becomes 1 2 3
Combinations possible from the remaining
1x1 1x2 1x3 2x1 and 3x1
Other combinations does not meet the requirements
So the answer is 5/25 = 1/5
Attachments
IMG_20191120_211555.jpg [ 1015.5 KiB | Viewed 4004 times ]
