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dear all I am reviewing a probability problem but I am a bit unsure the solution. Here is the problem:
males has 50% of being infected by diseaseX
females has 20% probability of being infected by diseaseX
the probability of a baby born from one infected parent to be infected is 30%; what is the probability of a random baby be infected?
I am ok with the calculations for the following combinations:
P(baby infected|mother NOT infected, father NOT infected)= 0*(1-0.2)*(1-0.5)=0
P(baby infected|mother infected, father NOT infected)= 0.3*0.2*(1-0.5)=0.03;
P(baby infected|mother NOT infected, father infected)= 0.3*(1-0.2) * 0.5=0.12

But when it comes to the probability of the baby being infected given both the parents are infected I am a bit confused, my first approach would be:
P(baby infected|mother infected, father infected)= 0.2*0.5*0.3^2=0.009
But from the solution I have on the slides I see this last one combination should be calculated as 1- the probability of the baby “escaping” the infection from both the parents:
P(baby infected|mother infected, father infected)= 0.2*0.5*(1-((1-0.3)*(1-0.3)))=0.05
It makes sense but what I am missing is the theoretical reason of why this is the only solution in this last case. I think it has to do with independence of events but I can’t really figure it out.
Any simple explanation would be highly appreciated!!
Thanks!!

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I think this comes down to the wording and interpretation of the setup. Perhaps a better wording would be "the probability of an infected parent passing the infection to their child is 30%." Then in order for a child of two infected parents to become infected, at least one of their two parents must pass the infection on to their child. The complement of this event is the event that both parents fail to pass on the infection, which happens with probability $(1-0.3)(1-0.3)=0.49$. Hence the probability that at least one parent passes on the infection is $1-0.49=0.51$.

If you instead calculate $0.3^2=0.09$, that is the probability that both parents pass on the infection. Certainly this would lead to an infected child, but that is not the only way for the child to be infected. It could also happen that the first parent passes on the infection and the second doesn't, which happens with probability $(0.3)(1-0.3)=0.21$. Or it could happen that the first parent fails to pass on the infection while the second parent does pass it on, which happens with probability $(1-0.3)(0.7)=0.21$. Notice if we add up the probabilities of these three cases we get $0.09+0.21+0.21=0.51$, which is the correct probability.

One final note is that the probabilities you've calculated are not conditional probabilities but rather probabilities of the intersection of the events. For instance, $$P(\text{baby infected} | \text{mother infected, father not infected}) =0.3$$ while \begin{align*} &P(\text{baby infected and mother infected and father not infected})\\ &=P(\text{baby infected} | \text{mother infected,father not infected})\cdot P(\text{mother infected, father not infected})\\ &=(0.3)\cdot(0.2)\cdot(0.5). \end{align*}

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  • $\begingroup$ thanks kccu that was helpful!! $\endgroup$ – Sergio_L9 Feb 20 at 9:23

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