Question

A man and woman have five children. what is the probability of this couple having four unaffected (either hbahba or hbahbs ) and one affected child if both parents are carriers?

Answer 1

The probability of the couple having four unaffected and one affected child if both parents are carriers will be A. 0.40.

How to calculate the probability?

The area of mathematics known as probability deals with numerical representations of the likelihood that an event will occur or that a statement is true. An event's probability is a number between 0 and 1, where, roughly speaking, 0 denotes the event's impossibility and 1 denotes certainty.

Unaffected = ¾; affected = ¼

The probability of this couple having four unaffected (either HbAHbA or HbAHbS ) and one affected = n!/a!b! (s^a) (t^b)

n= Total no. of progeny = 5

a = No. of unaffected = 4

b = No. of affected = 1

s = probability of unaffected = 3/4

t = probability of affected = 1/4

Therefore, the probability of the couple having four unaffected and one affected child if both parents are carriers will be:.

= 5! / 4!1! (3/4^4) (1/4^1)

= 120/24 (0.32) (0.25)

= 0.4 = 40%

Therefore, the probability is 0.4.

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A man and woman have five children. What is the probability of this couple having four unaffected (either HbAHbA or HbAHbS ) and one affected child if both parents are carriers?

a. 0.40

b. 0.25

c. 0.08

d. 0.66