Question

Question 5

Cystic fibrosis (CF) is a recessive gene disease meaning that a child has to inherit a defective CF gene from each parent. This means that there is a 25% chance of a child having cystic fibrosis if both parents are CF carriers Hence the number of children with CF (in a family whose parents are CF carriers) has a binomial distribution with n = the number of children in the family and p= 0.25

i. In a family of four (4) children whose parents are CF carriers, what is the probability that exactly one (1) of the children has cystic fibrosis. (2 mark)

ii. In a family of six (6) children whose parents are CF
carriers, what is the probability that one or more of the children
has cystic fibrosis.

Answer #1

i)

Sample size , n = 4

Probability of an event of interest, p = 0.25

**P ( X = 1) = C (4,1) * 0.25^1 * ( 1 - 0.25)^3=
0.4219**

...........

ii)

Sample size , n = 6

Probability of an event of interest, p = 0.25

X | P(X) |

0 | 0.1780 |

1 | 0.3560 |

2 | 0.2966 |

3 | 0.1318 |

4 | 0.0330 |

5 | 0.0044 |

6 | 0.0002 |

p(x>=1) = 1-p(0)

=1-0.178

**=0.8220**

................

THANKS

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Question 5
Cystic fibrosis (CF) is a recessive gene disease meaning that a
child has to inherit a defective CF gene from each parent. This
means that there is a 25% chance of a child having cystic fibrosis
if both parents are CF carriers Hence the number of children with
CF (in a family whose parents are CF carriers) has a binomial
distribution with n = the number of children in the family and p=
0.25
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