Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, the probability that their offspring will develop the disease is approximately 0.25. Suppose that a husband and wife are both carriers and that they have six children. If the outcomes of the six pregnancies are mutually independent, what are the probabilities of the following events? (Round your answers to four decimal places.)

a) What is the probability of all 6 children developing the disease? 0.0002

b) Only one child will develop the disease? 0.3559

c) The sixth child will develop the disease given that the first 5 didn't. 0.0593

Step-by-step explanation:

Hello!

The variable of interest is X: number of children of a couple that carries the gens, that develop Tay Sachs disease out of 6 children. The outcomes of the six pregnancies are independent.

The probability of a child developing Tay Sachs disease when both parents carry the genes to it is 0.25.

Symbolically P(S)= 0.25

Then the probability of the child not developing it is P(H)= 1 - 0.25= 0.75

a) What is the probability of all 6 children developing the disease?

If the 6 children develop the disease then you can describe the situation as:

"The 1st child is sick" and "The 2nd child is sick" and "The 3rd child is sick" and "The 4rth child is sick" and "The 5th child is sick" and "The 6th child is sick", symbolically:

P(S₁∩S₂∩S₃∩s₄∩S₅∩S₆)= 0.25*0.25*0.25*0.25*0.25*0.25= 0.25⁶= 0.0002

b) Only one child will develop the disease?

If only one is sick but you don't know which one is then you have the following possible scenarios:

"The 1st child is sick" and "The 2nd child is healthy" and "The 3rd child is healthy" and "The 4rth child is healthy" and "The 5th child is healthy" and "The 6th child is healthy"

or

"The 1st child is healthy" and "The 2nd child is sick" and "The 3rd child is healthy" and "The 4rth child is healthy" and "The 5th child is healthy" and "The 6th child is healthy"

or

"The 1st child is healthy" and "The 2nd child is healthy" and "The 3rd child is sick" and "The 4rth child is healthy" and "The 5th child is healthy" and "The 6th child is healthy"

or

"The 1st child is healthy" and "The 2nd child is healthy" and "The 3rd child is healthy" and "The 4rth child is sick" and "The 5th child is healthy" and "The 6th child is healthy"

or

"The 1st child is healthy" and "The 2nd child is healthy" and "The 3rd child is healthy" and "The 4rth child is healthy" and "The 5th child is sick" and "The 6th child is healthy"

or

"The 1st child is healthy" and "The 2nd child is healthy" and "The 3rd child is healthy" and "The 4rth child is healthy" and "The 5th child is healthy" and "The 6th child is sick"

Symbolically:

P((S₁∩H₂∩H₃∩H₄∩H₅∩H₆)∪(H₁∩S₂∩H₃∩H₄∩H₅∩H₆)∪(H₁∩H₂∩S₃∩H₄∩H₅∩H₆)∪(H₁∩H₂∩H₃∩S₄∩H₅∩H₆)∪(H₁∩H₂∩H₃∩H₄∩S₅∩H₆)∪(H₁∩H₂∩H₃∩H₄∩H₅∩S₆))=

(0.25*0.75*0.75*0.75*0.75*0.75)+(0.75*0.25*0.75*0.75*0.75*0.75)+(0.75*0.75*0.25*0.75*0.75*0.75)+(0.75*0.75*0.75*0.25*0.75*0.75)+(0.75*0.75*0.75*0.75*0.25*0.75)+(0.75*0.75*0.75*0.75*0.75*0.25)=

6*(0.25*0.75⁵)= 0.3559

c) The sixth child will develop the disease given that the first 5 didn't.

P(H₁∩H₂∩H₃∩H₄∩H₅∩S₆)= 0.*75*0.75*0.75*0.75*0.75*0.25= 0.25*0.75⁵= 0.0593

I hope this helps!

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