What is the probability that the one 3-digit and one 2-digit integers that could be formed out of 1, 2, 3, 4 and 5 (each figure is used for only once) are both even integers?
Give your answer as a fraction .
Is the answer 1/40?
1/10
Out all possible arrangements of the five digits, we are looking for:
OOEOE, and the spit the five digit into 3 + 2
Probability for OOEOE:
\frac{3}{5} * \frac{2}{4} * \frac{2}{3} * \frac{1}{2} * \frac{1}{1} = \frac{1}{10}
Still confused, why you consider it 5 digit, are not they separate integer consisting of 3 and 2 digits?
Please correct me if i am wrong
No of ways to form 3 digit even no = 4x3x2= 24 ways
Total 3 digit can form = 5x4x3= 60 ways
For two digit even integer = 4x2= 8 ways
Total 2 digit integer can form = 5x4= 20 ways
Probability = (24/60) x (8/20) = 2/5x2/5 = 4/25
The problem with your method is that there is repetition of digits between your 3 digit and 2 digit numbers
for example, 352 and 24 will not be an allowed number pair
That is why, I treat it like 5 digit number split in the middle
So that the number that are used for the first three digits do not occur in the last 2
Thanks, got it, need to read carefully