12. If a family is chosen at random from the set of all families with exactly three children, find the probability that: (2Marks)
a) the family has three boys if it is known that one child is a boy.
b) the family has three boys if it is known that the first child is a boy.

(a) 0.25

(b) 0.125

Step-by-step explanation:

The probability of the child being a boy or a girl is 50%.

That is, P (B) = 0.50 = P (G).

It is provided that the family selected has three children.

(a)

Compute the probability that the family has three boys if it is known that one child is a boy as follows:

Thus, the probability that the family has three boys if it is known that one child is a boy is 0.25.

(b)

Compute the probability that the family has three boys if it is known that the first child is a boy as follows:

The gender of the children are independent of each other.

So, if the first child is a boy then the probability of having three boys is:

P (3 boys | 1st boy) = P (1st boy) × P (2nd boy) × P (3rd boy)

                               = (0.50)³

                               = 0.125

Thus, the probability that the family has three boys if it is known that the first child is a boy is 0.125.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116