Question

A family has three children. Using b to stand for a boy and g for girl give the following.(To indicate that the first child is a boy, the second child is a boy, and the third child is a girl enter bbg. Enter your answers as a comma-separated list.). Let the event E that the family has exactly two boys.

a) Find the probability p(E)
b) Find the odd o(E) Part a): [Select ] Part b): ISelect ) ?

Answer 1

Answer:

3/8 and 3/5

Step-by-step explanation:

Given that a family has three children. Using b to stand for a boy and g for girl give the following we can write sample space for a family having three children as

(bbb, bbg, bgg, bgb, gbb, gbg, ggb, ggg)

E - exactly two boys

Sample space count = 8

Exactly two boys count = 3

a) p(E) = $\frac{3}{8}$

b) odds for E is probability/1-probability

= $\frac{3}{5}$

Answer 2

The probability helps us to know the chances of an event occurring. The probability that exactly two boys were born is 37.5%.

What is Probability?

The probability helps us to know the chances of an event occurring.

$\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}$

A.) As it is given that the family has three children, and each of the children can either be a boy or a girl, therefore, the number of possible options can be written as (bbb, bbg, bgg, bgb, gbb, gbg, ggb, ggg).

Now, if we look at all of the options the only options in which the number of boys is exactly two is 3 (bbg, bgb, gbb).

Further, the probability can be written as

$\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}$

$\rm Probability=\dfrac{\text{Number of options in which there are exactly two boys}}{\text{Total number of Possible options}}\\\\Probability(Exactly two boys) = \dfrac{3}{8} \\\\p(E)= 0.375 = 37.5\%$

Hence, The probability that exactly two boys were born is 37.5%.

B.) The number of possible options in which the number of boys is odd is 4 (bbb, bgg, gbg, ggb).

Therefore, the probability of having the odd number of boys can be written as:

$\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}$

$\rm Probability=\dfrac{\text{Number of options in which there are odd number of boys}}{\text{Total number of Possible options}}\\\\Probability(\text{odd number of boys}) = \dfrac{4}{8} \\\\Probability(\text{odd number of boys})= 0.50 = 50\%$

Hence, the probability of having the odd number of boys is 50%.

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