Probability Calculations for Random Samples and Genetic Inheritance

[brunettegurl]

1) A random sample of 2 people are selected from a group of 15 girls and 8 boys to participate in a study. Find the probability that the sample consists of 1 boy and 1 girl.

My take : n=23 P(1girl)*P(1boy)
(15/23)*(8/23)
= 0.2268
the answer should be 0.4743

2)Suppose that a disease is inherited via an autosomal recessive mode of inheritence.The children in the family have a probability of 1/4 of inheriting the disease. In a family of 4 children, find the probability that at least one of the children inherits the disease.

My take: n=4 x=1 p=1/4
nCx= n!/x!(n-x)! f(x)= nCx* p^x(1-p)^n-x

I get 0.4219
The answer should be 0.6836

Any help would be appreciated.

Thanks

 

[The Investor]

The simplest way to solve these problems:

1) P (1 boy 1 girl) = 1-P(2 girls) - P(2 boys)

2) P(one child or more inherits disease) = 1-P(no child inherits disease)These are very easy to work out, and don't require the level of complexity of the formulas you are using.

 

[brunettegurl]

Thanks!

 

[mathman]

If you want to do the first one directly, you need to do it properly.
girl first boy second = (15/23)(8/22)
boy first girl second = (8/23)(15/22)

Final answer (2x8x15)/(23x22)

 

[blue_raver22]

for providing these probability calculations for random samples and genetic inheritance. I would like to offer some clarification and corrections on the calculations provided.

1) For the first question, the probability of selecting 1 boy and 1 girl from a sample of 23 individuals can be calculated using the combination formula: nCr = n!/(r!(n-r)!). In this case, n = 23 and r = 2, so the calculation would be 23C2 = 23!/(2!(23-2)!) = 253. This represents the total number of possible combinations of 2 individuals that can be selected from a sample of 23.

Now, out of these 253 combinations, we want to find the specific combination of 1 boy and 1 girl. Since there are 15 girls and 8 boys in the sample, the probability of selecting 1 girl and 1 boy would be (15/23)*(8/23) = 0.2268, as you correctly calculated.

However, this is not the final probability as it represents only one possible combination out of the 253. To find the probability of selecting 1 boy and 1 girl in any combination, we need to divide 0.2268 by 253, which gives us the final probability of 0.0089.

2) For the second question, the probability of at least one child inheriting the disease can be calculated using the binomial distribution formula: P(x>=1) = 1 - P(x=0) = 1 - (nCx)*px*(1-p)n-x. In this case, n = 4 (number of children), x = 1 (at least one child inherits the disease), p = 1/4 (probability of inheriting the disease).

Substituting these values in the formula, we get P(x>=1) = 1 - (4C1)*(1/4)*(3/4)^3 = 1 - 4*(1/4)*(3/4)^3 = 1 - 0.4219 = 0.5781. This represents the probability of at least one child inheriting the disease in a family of 4 children.

I hope this helps clarify the calculations and provides the correct answers for these probability scenarios. Keep up the good work in your studies!