Difficult Math Problem #81 - Probability

My ans. is 1/4C2 = 1/6 (one pair will do the job (2,6) if of course we assume that we cannot pick same number twice.

1/3?

THE ONLY WAY WHEN SUM IS 8 IS WEN THE 2 NUMBERS ARE 2, 6
SO IT SHOULD BE 1/10 x 2/9= 1/45

There are 4 even numbers. So the all the possible set is 1/4 * 1/4.
The only way to 2 get sum of 8 is (2,6) and (6,2). Here these are 2 different set and should be counted twice.
So that answer should be 2/16 = 1/8.

There are 4 even integers: 2,4,6,8.
The probability to pick 2 or 6 = 1/4 + 1/4 = 1/2
Then the probability to pick the other integer (either 2 or 6) is 1/3
So the answer is 1/2 * 1/3 = 1/6

what's the OA? THere are like 5 different answers here

here's the OA:

Initially you have 4 even numbers (2,4,6,8 )
you can get the sum of 8 in two ways => 2 + 6 or 6 + 2
so the first time you pick a number you can pick either 2 or 8 - a total of 2 choices out of 8 => 1/2
after you have picked your first number and since you have already picked 1 number you are left with only 2 options => either (4,6,8 ) or (2,4,8 ) and you have to pick either 6 from the first set or 2 from the second and the probability of this is 1/3. Since these two events have to happen together we multiply them. ½ * 1/3 = 1/6

that google smiley 8) looks great
nice explanation !

dear 800guy
would you provide explanation for question no. 71 too?

You have four choices for even numbers (2, 4, 6 and 8)

Two of these numbers need to add to 8. Only one combination allows that (2 and 6).

There are two different ways to select 2 and 6. Either select 2 in your first attempt and select 6 in your second or vise versa. The probability will be the same.

Lets take the first scenario and figure out the probability.

First for us to get a 2 from 4 numbers (2,4,6 and 8), the probability is 1/4

Now we have 3 numbers left (4, 6 and 8). To get a 6 from these 3 numbers, the probability is 1/3

Multiply the two probabilities:

1/4*1/3 = 1/12

Now since the probability of the second scenario (where we pick 6 first and 2 second) is the same, you can add the two probabilities together:

1/12 + 1/12 = 2/12 = 1/6

HERE THE EVENT SUM =8 CAN BE OBTAINED BY THE COMBINATION OF THE FOLLOWING EVENTS:-
EVENT=2 AND EVENT =6 OR EVENT =6 AND EVENT =2
P(SUM=8) = 1/4*1/3+1/4*1/3 = 1/12+1/12 = 1/6

2 4 6 8 are even numbers between 1 and 9

2+6 is the required sum


so, 1/4C2=1/6

(2,4,6,8) are the even numbers betwen 1 and 9.
we can choose a sum of 8 in 2 ways, a) 2+6 OR b)6+2.
a)
1) Probability of getting a '2' out of 4 numbers = 1/4
2) Probabiltiy of getting a '6' out of 3 numbers = 1/3
Since we want both (1) AND (2) , 1/4*1/3 = 1/12

b)
The same values result as above , i.e., 1/12

Since our requirement is either a) or b) 1/12 + 1/12 = 2/12 = 1/6.

We can get the sum in two ways 2+6 or 6+2.
Total outcomes = 4C2

So probability of getting a 8 as the sum = 2/4C2 = 2/6 = 1/3

Where am i going wrong ? ? Can someone explain me ?

Should we use permutation here while calculating the total outcomes as 2+6 and 6+2 are considered two different options ? ?

sam2304 wrote:We can get the sum in two ways 2+6 or 6+2.
Total outcomes = 4C2

So probability of getting a 8 as the sum = 2/4C2 = 2/6 = 1/3

Where am i going wrong ? ? Can someone explain me ?

Should we use permutation here while calculating the total outcomes as 2+6 and 6+2 are considered two different options ? ?

Hi sam2304,

Let me review a few basics of Permutations & Combinations for you before I explain your doubt.

(i) Number of ways of arranging/distributing 'n' different objects on 'n' places is given by n!

(ii) Number of ways of selecting/choosing/picking 'r' objects from 'n' different objects is given by nCr. The 'C' stands for 'Combinations'.
For example:
If there are 4 people: A, B, C and D, the following are the various ways of choosing a 2-member team out of them.
(A, B), (B, C), (C, D), (A, C), (B, D), (A, D). There are 4C2 = 6 such teams. Notice that we are just concerned with the members in the team and not with arranging them.

(iii) Number of ways of arranging/distributing 'n' different objects on 'r' places (where r < n) is given by nPr. The 'P' stands for 'Permutations'.
For example:
If there are 4 people: A, B, C and D, the following are the various ways of distributing Ist and IInd prizes to 2 of them:
(A 1st, B 2nd), (B 1st, A 2nd), (B 1st, C 2nd), (C 1st, B 2nd), (C 1st, D 2nd), (D 1st, C 2nd) (A 1st, C 2nd), (C 1st, A 2nd), (B 1st, D 2nd), (D 1st, B 2nd), (A 1st, D 2nd), (D 1st, A 2nd). There are 4P2 = 12 such ways. Notice that we were concerned with arranging the two people.

(iv) OR is Addition, AND is addition.

Alright then, Do you see your mistake now?

The problem uses the word 'chosen', which indicates that we are not concerned with the arrangement. The 4C2, which is the total number of outcomes, is the number of ways of selecting 2 even numbers from 4 even numbers. It includes these: (2,4), (4,6), (6,8), (2,6), (4,8), (2,8).

How many of these add up to 8?

Just one: (2,6).

Required Probability = 1/6

An interesting Permutations & Combinations concept:
https://www.beatthegmat.com/important-pe ... 11283.html