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Two numbers are independently selected from the set of positive integers less than or equal to \(6\). What is the probability that the sum of the two numbers is less than their product? Express your answer as a common fraction.

 Apr 14, 2022
 #1
avatar+2437 
+1

There are \({6 \choose 2 } = 15\) ways to choose 2 numbers. 

 

The only way the sum will be less than their product is if neither of the digits is 1. 

 

There are \({5 \choose 2} = 10\) ways to choose 2 numbers without a 1.

 

Thus, the probability is \({10 \over 15} = \color{brown}\boxed {2 \over 3}\) 

 Apr 14, 2022
 #2
avatar+114 
-1

To find the probability, first find all possible cases (note that zero is neither positive or negative)

 

2 are picked out of 5, so

 

Using the combination formula : 

 

Which is - 

 

Now that is the number of possible cases of 2 numbers getting chosen out of 5.

 

 

Now lets list all the cases of "the sum of the two numbers is greater than their product" (note that one number can't be chosen twice)

 

1 and 2

 

1 and 3

 

1 and 4

 

1 and 5

 

And thats all (4 cases)

 

so the answer is 

 

2/5

 Apr 14, 2022
 #3
avatar+2437 
0

Read carefully, x is "less than or equal to 6"...

BuilderBoi  Apr 14, 2022
 #4
avatar+124594 
+1

Let's see if  we can do this by a table  ⇒   x,y  table entries   denote  sum, product

 

      1         2        3         4          5          6

 

1             3,2      4,2       5,4      6,5      7,6 

 

2                        5,6       6,8      7,10    8,12

 

3                                   7,12      8,15   9,18

 

4                                               9,20    10,24

 

5                                                          11, 30

 

6

 

 

We  have 15  possible outcomes  and  10  fit the criteria

 

So....the probability  =  10  /15   =  2 / 3

 

cool cool cool

 Apr 14, 2022

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