Let us calculate the inverse probability.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
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,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 ?
i got the same answer?abhishekg21 wrote:shouldn't it be 2/7.
total ways of selecting 2 cards out of 8 is =8*7=56 (considering the order)
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 ?
Your approach is fine.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.
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.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
