Useful addition rules for mutually exclusive events: Rule 1: Given N mutually ex

Question

Useful addition rules for mutually exclusive events:

Rule 1: Given N mutually exclusive events {A1,A2, ... ,AN} then

P(A1) + P(A2) + ... + P(AN) = 1.0

Rule 2: P(A particular mutually exclusive event) = 1 - P(all the others) This rule is very useful!

For example:

Suppose I toss a coin 10 times (or 10 coins once) and ask what is the probability of getting at least one head in 10 tosses. We could do this in two ways one is to sum probabilities P(1 at least head in 10) + P(2 heads in 10) + P(10 heads in 10) or we can simply go:

P(1 at least head in 10 tosses) = 1 - P(no heads in 10 tosses) = 1 - 1/(2^10)

Application to genetics:

Activity 6

For maternal and paternal chromosomes in the human gamete example, what is the probability of a gamete having at least one maternal chromosome?

Activity 7

Many types of color blindness are what are called X linked, that is determined by genes on the X chromosome. Suppose a woman is carrying one X chromosome with the gene for a particular type of color blindness; her other X chromosome does not have this gene. If she is married to a man who does not have this gene on his X chromosome. You may remember that color blindness is X linked recessive.

A. What is the probability that her first child will carry the X chromosome with the gene associated with color blindness?

B. Suppose amniocentesis reveals that the child is male. What is the probability that the child is color blind. Hint: this involves conditional probability

Explanation / Answer

6. Each gamete carries either the maternal or paternal chromosome, so a single pair of chromosomes has two possibilities of getting either the maternal one or paternal one. Humans have 23 pairs of homologues chromosomes, means there are 2^23 possible combinations of maternal and paternal chromosomes in each gamete.

Thus, the probability of a gamete to have at least one maternal chromosome = 1- P (having no maternal chromosomes) = 1- 1/ 2^23

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