If three prime number are randomly selected from prime numbers less

To have the sum as an even number one of the numbers should be even.

ODD+ODD+EVEN=EVEN
ODD+EVEN+ODD=EVEN

Logically we can conclude that the probability that the prime number 2 will be selected out of 9 other odd numbers is relatively low.

total possibilities=10*9*8=720
total odd possibilities=9*8*7=504
total even possibilities=720-504=216
216/720=.3=30%
C

pr3mk5: Your answer is brilliantly short and intuitive - thanks. It contained the following:
"Even prime can be selected in 1C1 ways = 1
Odd prime can be selected in 9C2 ways = 36

Total ways in which 3 numbers can be selected = 10C3 = 120"

Could you please elaborate more on how to get the "total ways" - > didnt get the "10C3" (actually dont know what's the "C" in there)

Thanks

There are 10 prime numbers that are less than 30:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Only 2 is even , rest are odd

Now if the sum of three prime numbers is even , 2 must be selected in one of the three number otherwise the result will always be ODD as ODD +ODD+ODD = ODD

Now there are three ways -
1) 2 AND any odd number AND any odd number
OR
2) any odd number AND 2 AND any odd number
OR
3) any odd number AND any odd number AND 2

Probability -
(1/10)*(9/9)*(8/8)
+
(9/10)*(1/9)*(8/8)
+
(9/10)*(8/9)*(1/8)
= 3/10=0.3

Ans C

29, 23, 19, 17, 13, 11, 7, 5, 2

Total number of ways to choose any three numbers from nine \(9C3 = \frac{9*8*7}{3!(9-3)} = 84\)

Two odd numbers to choose from 9 numbers: \(9C2 = \frac{9*8}{2!(9-2)} = 36\)

Now we are left with 7 numbers so there is only one even number hence I can choose only one number from 7 ---- >

\(7C1 = \frac{7!}{(7-1)}= 7\)

\(Two Odd + One Even = 9C2 +7C1 = 36+7 =43\)

So probability is \(\frac{43}{84}\) = ca. \(0.5\)


hello there pushpitkc - what, where and why ? it looks like a logicaly "correct" solution but i got incorrect answer what did i do wrong

Hey dave13

You did everything right except for this - there are 10 prime numbers less than 30.

The total possibilities will change accordingly. This is what the correct solution would look like

Listing all primes less than 30 - 29,23,19,17,13,11,7,5,3,2

Total ways of selecting 3 numbers from 10:\(C_3^{10} = \frac{10*9*8}{3*2*1} = 120\)

There is only one possibility to choose 3 prime numbers and have an even sum,
when we have 2 odd and 1 even prime number as the 3 primes need to be different

Total possibility of choosing 2 odd primes from 9: \(C_3^{9} = \frac{9*8}{2*1} = 36\)

There is only one possibility of choosing an even prime number. Therefore, Probability is \(\frac{36}{120} = 30\)%(Option C)

Hope this clears your confusion!

judsonvpires
C stands for combinations

\(nCk = \frac{n!}{(n-k)!*k!}\)

\(10C3 = \frac{10!}{(10-3)!*3!} =\frac{10!}{7!*3!}\)

The simple logic here is to understand that in reality there are two cases here. As we have only one even number in the list of prime numbers (only 2), there can be only two cases to choose three prime numbers:
1st case: odd+odd+odd
2nd case: odd+odd+even
So, to make our calculations easier, we can subtract the 1st case from the total or 1 to get the 2nd case.

In order to calculate the possibilities of the 1st case, we should do a simple probability calculation. As there are ten prime numbers in that range: 2,3,5,7,11,13,17,19,23,29.
9/10*8/9*7/8=504/720
And finally subtract 504/720 from the total, 1 to get 0.3, which in percent form is 30%.

There are 10 prime numbers below 30 and only one such number is even. For the sum of three selected numbers to be even, we have to select the even prime number (i.e. 2). If we don't select it, the sum will be odd. And if we select it, doesn't matter which other numbers we choose, the sum will be even.

So, essentially, the probability asked in the question is the same as the probability that we'll select 2 as one of the three numbers.

So, we have to select 2 as one of the numbers and the other two numbers can be any of the other 9 prime numbers. So, total ways to select such a set of three numbers out of a set of 10 prime numbers is 1C1 * 9C2.

Total number of ways to select 3 numbers out of 10 numbers is 10C3.

The required probability is \(\frac{9C2}{10C3}\) i.e. 0.3.

I hope it helps.

- CJ

The prime numbers less than 30 are:

2, 3, 5, 7, 11, 13, 17, 19, 23 and 29

The number of ways of picking any 3 of them is 10C3 = 10!/(3! x 7!) = (10 x 9 x 8)/(3 x 2) = 120.

Our goal is to add 3 primes to obtain an even sum. Note that if all 3 prime numbers are odd, we will never obtain an even sum, so we know that one of the 3 primes chosen must be the number 2. So, from the remaining 9 primes, we need to pick any two of them. The number of ways of picking 2 primes from 9 is 9C2 = 9!/(2! x 7!) = (9 x 8)/2 = 36. Therefore, the probability is 36/120 = 3/10 = 30%.

Answer: C

Problem can be solved within 2mins.

Solution :

Now for sum to be odd, we need OOE because Odd+Odd+Even = Even , since Odd+Odd = even then Even+Even = Even

Now the OOE can EOO or OOE it doesnt matters so 3!/2!

Now prime numbers <30 are only 10 , of which only 2 is even . So 2 can be selected in 1/10 and other ODD Numbers can be in 9/9 and 8/8

result : 1/10*9/9*8/8

since we removed duplicate variations so final will be 3!/2! * 1/10*9/9*8/8
= 3/10 = .3 = 30%

When using Counting Methods to answer Probability questions, one is always relying on the basic definition of the probability of an event’s occurrence:

(Count of Favorable Outcome)
__________________________
(Count of Total Possible Outcomes)

The 1st step is always to define WHAT that “Favorable Outcome” entails.

For this problem, the favorable outcome is the following:

From the set of the prime numbers less than 30, we choose 3 of those numbers and the SUM is Even.

A good tip to remember whenever dealing with quantitative aptitude questions involving Prime numbers is that 2 is the only Even Prime number. More often than not, this key insight is the breakthrough needed to answer the question.

Set of Primes less than 30:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]

Since we are choosing 3, distinct prime numbers from this set, the total possible number of ways in which we can choose is found by:

“10 choose 3” = #no. ways we can make groups of 3 distinct numbers from out of 10 numbers = (Count of Total Possible Outcomes) = Denominator

(10!) / (7! 3!) = (10)(3)(4)

Next, since 9 of the 10 primes are ODD and only the number (2) is EVEN, we can see that the only way to get a Favorable Outcome is when we have:

(Even) + (Odd) + (Odd)

In other words, every favorable outcome MUST include the element (2) from the set.

The next two numbers chosen from the set do not matter. We can count the number of ways to group 2 numbers out of the remaining 9:

(9!) / (7! 2!) = (9)(4)

Probability = (Count of these Favorable Outcomes) / (Count of total possible outcomes)

= (9)(4)
________
(10)(3)(4)

= 3/10 = 30%

Answer: 30%

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