The possible outcomes that satisfy the condition of "at least one house gets the wrong package" are:
One house gets the wrong package or two houses get the wrong package or three houses get the wrong package.
We can calculate each of these cases and then add them together, or approach this problem from a different angle.
The only case which is left out of the condition is the case where no wrong packages are delivered.
If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above.
There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package.
The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.
1×1×1=1
Determine the total number of ways the three packages can be delivered.
3×2×1=6
The number of ways at least one house gets the wrong package is:
6?1=5
Therefore there are 5 ways for at least one house to get the wrong package.
4th observation i.e., 8 is the median.
47/10000 = .0047
.02 =(2/100) x 100% =2%
Let N / 11 = 233
Then, N = 233 x 11 = 2563
? Missing digit is 5.
121012 = 12 x 10084 + 4
? remainder = 4
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.