How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits a
1 answer:
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a) Probability of both being males is 27%
b) Probability of both being females is 23%
c) Probability of having exactly one male and one female is 50%
Step-by-step explanation:
a)
The probability that the birth is a male can be written as
(which corresponds to 52%)
While the probability that the birth is a female can be written as
(which corresponds to 48%)
Here we want to calculate the probability that over 2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:
And substituting, we find
So, 27%.
b)
In this case, we want to find the probability that both children are female, so the probability
As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:
And substituting
We find:
Which means 23%.
c)
In this case, we want to find the probability they have exactly one male and exactly one female child. This is given by the sum of two probabilities:
- The probability that 1st child is a male and 2nd child is a female, namely
- The probability that 1st child is a female and 2nd child is a male, namely
So, this probability is
We have:
Therefore, this probability is
So, 50%.
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