pzazz12 wrote:All envelope contains eight bills:2 ones,2 fives ,2 tens,and 2 twenties.Two bills are drawn at random without replacement .what is the probability that their sum is $20 or more?

A.1/4
B.2/7
C.3/7
D.1/2
E.2/3

Let us calculate the inverse probability.
If the sum of two bills is less than $20, the bills must be a pair from the set {$1, $1, $5, $5, $10, $10} with a reduced pair ($10, $10).

Total number of selection of 2 bills from 6 bills = 6C2 = 15.
Total number of selection of 2 bills from 8 bills = 8C2 = 28.
Probability that the sum of the selected two bills is less than $20 = (15-1)/28 = 1/2

Therefore, probability that the sum of the selected two bills is $20 or more= 1 - 1/2 = 1/2.

The correct answer is D.

Yes we can consider the order, but in that case also the result will be same. Because 2 will be multiplied both in the numerator and denominator.

Flaws in your procedure are in the second part.

abhishekg21 wrote: ...out of options availbale(1,1,5,5,10,10,20,20) we can have 20 or more in following ways

(1,20),(5,20),(10,20),(20,20),(10,10)

now ways of selecting the above options considering the order
(1,20) in =2*2 =4
(5,20) =2*2=4
(10,20)=2*2=4
(20,20)=2*1=2
(10,10)=2*1=2
so prob is =16/56=2/7

please tell me what is wrong in this approach ?

Considering the order (1,20), (5,20) and (10,20) can be selected in 8 ways not in 4 ways as there are 2 of each of them. Say for the pair (1,20), there are two $1 bill 1a and 1b, and two $20 bill 20a and 20b. Now considering the order, two of them can be selected in the following ways,
(1a,20a), (1a,20b), (1b,20a), (1b,20b), (20a,1a), (20a,1b), (20b,1a) and (20b,1b).

Therefore,
(1,20) in = 2*2*2 = 8 ways
(5,20) = 2*2*2 = 8 ways
(10,20)= 2*2*2 = 8 ways
(20,20)= 2*1 = 2 ways
(10,10)= 2*1 = 2 ways

So probability = (8+8+8+2+2)/56 = 28/56 = 1/2.

aspirant_mumbai wrote:Hi All,

Here is my approach. I am using slot method.

We have two slots. Lets fix first one for 20, & the second slot for rest of 7 numbers.
So the total no of ways of selecting a sum greater than 20 is 1*7=7
Again repeating the same fixing 10 at slot one:
Total number of ways 1*3 (thats selecting 10,20,20) =3
Fixing 5 at slot one:
Total number of ways 1*2(selecting 20,20) = 2
For 1 at slot number 1
Total number of ways 1*2 (selecting 20,20) = 2

Total ways = 2*(7+3+2+2)=14*2

Total ways to selecting two numbers our of eight = 8*7
Probability = 14*2/8*7 = 1/2

Rahul Lakhani, please correct my approach if wrong.

Your approach is fine.

pzazz12 wrote:All envelope contains eight bills:2 ones,2 fives ,2 tens,and 2 twenties.Two bills are drawn at random without replacement .what is the probability that their sum is $20 or more?

A.1/4
B.2/7
C.3/7
D.1/2
E.2/3

The number of ways to select 2 bills from 8 bills is 8C2 = 8!/[2!(8-2)!] = (8 x 7)/2! = 28. Now we have to determine the number of ways to select 2 bills that total $20 or more.

If one of the two bills is one of the $20 bills, then the sum is guaranteed to be $20 or more. Thus, for each $20 bill selected, we can pair it with the 6 non-$20 bills. Since there are two $20 bills, we have 6 x 2 = 12 pairings. Furthermore, we have two more pairings: we can pair the two $20 bills with each other and we can also pair the two $10 bills with each other. Thus, we have 14 pairings of two bills that total $20 or more. Therefore, the probability is 14/28 = 1/2.

Answer: D