Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

14% of people are left handed, if we want to find the chances of a brother and a sister both being left handed, we might be tempted to multiply .14 by .14. What is wrong with this reasoning when trying to find out the probability of both of them being left handed?

share|cite|improve this question
Is the probability of a male being left handed, and a female being handed the same? – JavaMan Mar 21 '11 at 7:49
Why should it be wrong? – Did Mar 21 '11 at 7:54

The main problem is that the events that the brother and sister are left handed are not known to be independent (and I would expect them not to be independent if there are either genetic or environmental factors which influence handedness). It is only when you know that events $A$ and $B$ are independent that $P(A\cap B) = P(A)P(B)$.

Similarly, if $5\%$ of the population is named Smith, and the siblings are too young to have changed last names, then the chance they are both named Smith might be close to $5\%$ rather than $5\% \times 5\%$.

A more subtle issue is whether your handedness is independent of the number of siblings you have.

share|cite|improve this answer

You have to take into account the total number of people you took that 14% number from.

Lets say 50% of people are left handed. If you sampled that number from the brother and sister, the probability that both are left handed is 0%.

If the brother and sister are not part of the sample your reasoning would be correct.

We have 0.14=left-handed/total people=L/T

The correct cacluation would be P=L/T*(L-1)/(T-1). For a large sample P approaches (L/T)^2

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.