If x and y are each randomly chosen from the integers 1, 2, 3, 4, and

Question

If x and y are each randomly chosen from the integers 1, 2, 3, 4, and 5, what is the probability that the product xy will be greater than 10 ?

(A) 4/25
(B) 1/5
(C) 6//25
(D) 8/25
(E) 2/5

Bunuel · Accepted Answer

There are 5 options for x and 5 options for y, so the total number of ways to choose x and y is 5 * 5 = 25.

Out of these 25 pairs of (x, y), the following 8 give a product greater than 10:

x = 3 and y = 4
x = 4 and y = 3

x = 4 and y = 4

x = 5 and y = 3
x = 3 and y = 5

x = 5 and y = 4
x = 4 and y = 5

x = 5 and y = 5

The probability, therefore, is 8/25.

Answer: D.

celine_ng · Answer

x,y = 1,2,3,4,5

x*y > 10

Total number of cases will be 5*5 = 25

We can have 8 pairs as below: 
3*4, 4*3 
3*5, 5*3
4*5, 5*4
4*4 
5*5

The probability = 8/25 => D

Archit3110 · Answer

Quest has no condition of repetition not allowed so pairs possible are
5×5=25
3,4 
3,5
4,3
4,4
4,5
5,5
5,4
5,3
8 pairs
8/25
Option D

Posted from my mobile device

rey1009 · Answer

total possible products = 5*5

xy > 10 only when xy are : 3,4,5
3,4; 3,5;4,3;4,4;4,5;5,3;5,4;5,5 => 8

prob =8/25

rakman123 · Answer

Can someone explain why, when selecting 2 numbers of the 5 (when picking x & y), it does not become calculated as 5c2? Which gives us 10 instead of 25 (for the denominator)

Thank you

Anki111 · Answer

As per me, 5C2 is not used as it will not count the repetition number in a set like (1,1) (2,2) (3,3) etc. But in the question, it doesn't mention that we can't count repetition. Hence, all the numbers will be counted by 5*5. 
Similar, to the example of a dice. When 2 dice are rolled, total outcomes are 36

Confide · Answer

Almost got it incorrectly, realized the question does not mention anything about replacement. I assumed we can use the same numbers for x and y.

We have 25 cases and only 8 out of the 25 cases satisfy our condition for xy>10

Guntabulla · Answer

Can someone please explain what makes this a 800+ question?
It seems pretty straightforward

Fish181 · Answer

It's a trick problem. Not only will people under time pressure not consider the inverse of possible pairs, they also have to understand the implied repetition that is allowed.

nisen20 · Answer

this wording is terrible.
i bet it made at least half of people here fail to figure out test maker's intended meaning—x and y can be the same number.

1~5          1~5      there're 25 pairs.
     5            3~5      if 5 was chosen as one of the two numbers, another could be 3, 4 or 5.   {5, 3} {5, 4} {5, 5} 
     4            3~5      if 4 was chosen as one of the two numbers, another could be 3, 4 or 5.   {4, 3} {4, 4} {4, 5} 
     3            4~5      if 3 was chosen as one of the two numbers, another could only be 4 or 5.   {3, 4} {3, 5}

how many different pairs that generate a product greater than 10?  8.
p = 8/25

Oppenheimer1945 · Answer

2x({3,4} or {3,5} +{4,5}) +{4,4} +{5,5}/ {1,2,3,4,5}=(6+2)/25=8/25

Purnank · Answer

we all know for product to be greater than 10 we need numbers more than 2 as 10 = 5*2 
so there are 3 possible numbers for each x and y so total possiblities = 9 But in that 3*3 is not possible so reduce 1
Total outcome becomes = 9-1 = 8
and total number of sample space is = 5*5 = 25
Probability = 8/25

Answer is D.

__thisisjay · Answer

The question does not justify do we take all a one or we have 2 sets and we have to select from both

Bunuel · Answer

x and y are each randomly chosen from the integers 1, 2, 3, 4, and 5 means x and y are picked separately from the integers 1, 2, 3, 4, and 5. x is picked from 1, 2, 3, 4, 5, and y is picked from 1, 2, 3, 4, 5. This allows for both values to be the same or different, with no restriction on repetition.

demawtf · Answer

If x and y are each randomly chosen from the integers 1, 2, 3, 4, and 5, what is the probability that the product xy will be greater than 10 ?

(A) 4/25
(B) 1/5
(C) 6//25
(D) 8/25
(E) 2/5

1 case 
5*4=20 --> P:1/5*1/5 = 1/25 same for below so 2*1/25
4*5=20

2 case --> P: 2*1/25
3*4=12 
4*3=12

3 case --> P: 2*1/25
3*5=15 
5*3=15

4*4 --> P: 1/25
5*5 --> P: 1/25

Total P: 2/25*3 + 1/25*2 =8/25

mbaprepavi · Answer

Hi  Bunuel 
The question says "product of xy >10"
A fundamental question - although there are 8 ways to choose x and y
The product > 10 will only have 5 ways

Why is the answer not 5/25 i.e 1/5

mbaprepavi · Answer

Understood, i found the probability of the product being greater than 10
Therefore i took each value of the product uniquely and got 5 cases..
But its the ways to select x,y..

Thanks

KarishmaB · Answer

Responding to a pm: Think of this as throwing 2 five sided dice, each with numbers 1 to 5 on them. We want that the product on the two should be greater than 10. Since we are selecting x and y, the two dice are different say one is red and the other is blue. The results would look like (1, 1), (1, 2), (1, 3)... (2, 1), (2, 2) etc. Product is greater than 10 when 3 is multiplied by 4 or 5 or when 4 and 5 are multiplied in some way. x = 3, y = 4/5; Number of ways = 2 x = 4/5, y = 3; Number of ways = 2 x = 4/5, y = 4/5; Number of ways = 4 Total 8 favorable cases. Total number of cases = 5 * 5 = 25 Required Probability = 8/25 Answer (D) Why can we not use 3C2/5C2 ? Because the two number can be the same. When we select 2 out of 5, we are selecting 2 distinct numbers out of 5. But here, they need not be distinct. Here is another interesting question on Probability: https://anaprep.com/combinatorics-visua ... obability/

Dream009 · Answer

My approach

if X Y selection as below
x = 5 y can be slected 2 ways (no 10 is included)
x = 4 y is 2 ways
x = 3 y is 3ways
x = 2 y is 5 ways
x = 1 y is 5 ways 
Sum is 17 ways of under 10 or 10. So greator than 10 is 8 ways

If x and y are each randomly chosen from the integers 1, 2, 3, 4, and

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