Calculating Genetic Probabilities with Punnet Squares

  • Thread starter Thread starter at2341
  • Start date Start date
  • Tags Tags
    Probability
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 4K views
at2341
Messages
10
Reaction score
0
I'm trying to work out genetic probabilities using the punnet square.

Let's say I have 2 shirts and 2 pants. The odds of me wearing shirt #1 is 1/2. The odds of me wearing pants #1 is 1/2. Does this mean that the odds of me wearing shirt #1 or Pants #1 is 1 (100%) (addition rule - 1/2 + 1/2). This seems counterintuitive.

Is the answer that I have to subtract out from 100% the odds of me wearing BOTH shirt #1 and pants #1?

Thanks
 
Physics news on Phys.org
at2341 said:
I'm trying to work out genetic probabilities using the punnet square.

Let's say I have 2 shirts and 2 pants. The odds of me wearing shirt #1 is 1/2. The odds of me wearing pants #1 is 1/2. Does this mean that the odds of me wearing shirt #1 or Pants #1 is 1 (100%) (addition rule - 1/2 + 1/2). This seems counterintuitive.
No, it's not simple addition.

When you are looking at the probability of wearing shirt 1 OR pants 1, how many possible outcomes total do you have of shirts and/or pants? And how many possible outcomes would fulfill the requirement of having shirt 1 OR pants 1? [/quote]
Is the answer that I have to subtract out from 100% the odds of me wearing BOTH shirt #1 and pants #1?
No, that's not what you would subtract out. Though, one should be careful of how the question is phrased. If you need to determine any combination where shirt 1 or pants 1 (or both) is included, then both is within that same set of outcomes you need to consider, and the only outcome that will NOT satisfy your conditions is shirt 2 and pants 2 together.

On the other hand, if it says shirt 1 OR pants 1, but NOT both, then the outcome with shirt 1 AND pants 1, as well as the outcome of shirt 2 AND pant 2, are both excluded.