Two numbers are randomly selected from following set {3, 4,7,8,and 9}. What is the probability that

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05 Jan 2017 Probability

Two numbers are randomly selected from following set {3, 4,7,8,and 9}. What is the probability that product of these 2 selected numbers is an even number?

OPtion

1) 3/10
2) 7/20
3) 2/5
4) 7/10
5) 3/20
6) 1/8
7) 9/20
8) 11/20
9) 13/20
10)9/10

Solution

For faster solution let’s find out probability of Product of 2 number being odd
Product of 2 numbers a and b would be odd only if both a and b are odd
Odd numbers in given set {3,7,9}
Number of such numbers where product of 2 selected number is odd = 3C2 = 3*2/2 = 3
Total Number of possible ways to select 2 numbers from set {3,4,7,8,9} = 5C2 = 5*4/2 = 10
Probability that product of these 2 selected numbers is odd = 3/10
Probability that product of these 2 selected numbers is even = 1-
(3/10) = 7/10

Option 4)]

Correct Option: 4 Best Solution (16)

madanpal vishwakarma    8 years AGO

the possible ways to select 2 number and product
3 x 4 =12
3x7= 21
3x8 =24
3x9 =27

4x7 =28
4x8=32
4x9 =36

7x 8 =56
7x9= 63

8x9 =72


So clearly we have total 10 pair out of these 7 pair has even product value

so the probability is 7/10

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Mukund Manohar   8 years AGO

Total Possible Outcome
= 10, i .e . (3,4),(3,7),(3,8),(3,9),(4,7),(4,8),(4,9),(7,8),(7,9),(8,9)
Positive Outcome = 7, i .e .
(3*4)=12,(3*8)=24,(4*7)=28,(4*8)=32,(4*9)=36,(7*8)=56,(8*9)=72
Required Probability
= Positive Outcome/Total Possible Outcome
= 7/10

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Siva   8 years AGO

Total selection ways 5c2=20, required events are 14. (3,4),(3,8),(4,3),(4,7),,(4,8),(4,9),(7,4),(7,8),(8,3),(8,4),(8,7),(8,9),(9,4),(9,8). So the probability is 14/20=7/10

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Vivek   8 years AGO

Total combination from the given random numbers = 5C2 = 10
Combination of two numbers whose product will be even are = {3,4}, {3,8}, {4,7}, {4,8}, {4,9}, {7,8}, {8,9}
Total count = 7
Probability = 7 / 10
Option 4 is the correct answer.

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Pragati Rode   8 years AGO

Product of two numbers are even
={3,4},{3,8},{4,7},{4,8},{4,9},{7,8},{8,9}
=7
total = 5C2= 5!/3!*2! = 5*4/2 = 10

probability that Product of two numbers are even =7/10
so option 4

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Avinash   8 years AGO

product of two even no's is even and even and odd is also even
then take the combinations
probability= 7/5C2

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veera babu   8 years AGO

randomly means selected pair not repeated
{3,4,7,8,9}=5c2=10
product pairs={(3,4),(7,4)(9,4)(3,8)(7,8)(9,8)(4,8)
totally 7
7/5c2=7/10

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Himanshu Kumar   8 years AGO

product of two no. is even
no. 0f total outcomes are 5*4= 20
3,4 3,8 4,7 4,8 4,9 7,9 8,9 each one comes two times
7*2=14
14/20=7/10

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vaghani rushi   8 years AGO

total testcase are 5C2. and product of two number is even if and only if any one or both should be even. so total number for that product is even are 7 so 7/5C2 = 7/10

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sruthi pujari   8 years AGO

Total selections:5*4=20
In that odd*odd are to b rejected to get even product. (3,7),(7,3),(3,9),(9,3),(7,9),(9,7)
=> 20-6=14 pairs
Prob=14/20=7/10

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v kusuma   8 years AGO

There are 10 choices for selecting two numbers.(3,4),(3,7),(3,9),(4,7),(4,8),(4,9),(7,8),(7,9),(8,9).
the chance of product of two numbers is even is 7.
the answer is 7/10

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Lakshmi Priya   8 years AGO

probability of occurence of an event is 14
total no of possibilities is 20
14/20=7/10

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Pabitra Paul   8 years AGO

getting 2 number out of 5 will be happen in 10 ways.
products of odd number will be odd. so there are 3 odd numbers so the probability will be 3/10
now the product of two numbers will be even will be 1-(3/10)=7/10

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saif husain   8 years AGO

total products using two numbers are 10
out of which 7 are even

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Surya   8 years AGO

number of sample space is 10
ie (12,21,24,27,28,32,36,56,63,72)
then number of events is 7
ie (12,24,28,32,36,56,72)
so by rule it is 7/10
option 4
###########

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priyadharshini   8 years AGO

here the probability is given as 7/10.and no of samples is 5.so the no of favorites be 7 .therefore probability is 7/10

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