1. four digits 1,2,4 & 6 are selected at random to form a four digit number.what is the probability that the number so formed,would be divisible by 4? ans is 1/3,how?
2. a packet of 10 electronic components is known to include 3 defectives.if 4 components are selected from the packet at random,what is the expected value of the number of defective? ans is 1.20,how?
Answer of 1;
Solution,
let the probability of the no. so formed after dividing by 4 be denoted by event A
The total no of numbers that can be formed by 4 digits can be obtained through factorial of 4
4!=4*3*2*1=24
so, total likely events=24
Now, favourable outcomes can be obtained as
all forming 24 differents numbers with 4 digits 1,2,4,6 will be divisible by 4 if they have last two numbers that can be divided by 4
i.e. those numbers which have 12,16,24 and 64 at last will be divisible by 4.
for example 6412, 2416 and so on...
Similarly, 6412 and 4612 are different numbers so, favourable outcomes(m) would be 4*2=8
Lastly, P(A)=m/n
8/24=1/3
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