1.1.1 Multiplying probabilities

If you toss two coins, what is the probability that both coins are “heads”? Assume the two coins are described by probabilistic bits

a=(a0a1)andb=(b0b1),\displaystyle a=\begin{pmatrix}a_{0}\\ a_{1}\end{pmatrix}\qquad\text{and}\quad b=\begin{pmatrix}b_{0}\\ b_{1}\end{pmatrix}, (1.12)

where outcome 0 corresponds to “heads” and outcome 1 to “tails”. Then the probability to get “heads” for coin aa is a0a_{0} while for coin bb it is b0b_{0}. (We don’t assume that the coins are fair, so these probabilities are not necessarily 50%.) The probability that both coins simultaneously show “heads” is given by multiplying the probabilities of the two individual events:

p00=a0⁢b0.p_{00}=a_{0}b_{0}. (1.13)

Note that p00≤a0p_{00}\leq a_{0} and p00≤b0p_{00}\leq b_{0} since a0≤1a_{0}\leq 1 and b0≤1b_{0}\leq 1. This is intuitive, since getting “heads” simultaneously for both coins should be no more likely (and is typically less likely) than getting it for any of the coins individually. You can similarly compute the probabilities of all other combinations of heads and tails. We summarize all four cases in the following table:

p00\displaystyle p_{00} =a0⁢b0,p01=a0⁢b1,\displaystyle=a_{0}b_{0},\qquad p_{01}=a_{0}b_{1}, (1.14)
p10\displaystyle p_{10} =a1⁢b0,p11=a1⁢b1.\displaystyle=a_{1}b_{0},\qquad p_{11}=a_{1}b_{1}.

We call two events independent if they originate from two different sources, and the occurrence of one of them doesn’t tell you anything about the occurrence of the other. Typically such situation is described using the word “and”. For example, “the first coin is heads and the second coin is tails”. We multiply probabilities if we want to know if two independent events occurred simultaneously.

Exercise 1.2 (Multiplying probabilities).

Alice is bored during her math class and starts looking at her digital watch. The second’s counter on her watch can show values from 00 to 59. Assume that at some random point within the next minute Alice looks at the second’s counter on her watch.

  1. 1.
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    What is the probability that she sees 00?

  2. 2.
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    What is the probability that the last digit is 0?

  3. 3.
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    What is the probability that the first digit is 0?

  4. 4.
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    Argue that the values of both digits are independent from each other. Verify your answer to question 1 by multiplying the probabilities from questions 2 and 3.

Solution.
  1. 1.
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    The counter can have 6060 different values. The probability to see any of them is 160\frac{1}{60}.

  2. 2.
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    The last digit has 1010 different values. The probability to see any of them is 110\frac{1}{10}.

  3. 3.
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    The first digit has 66 different values. The probability to see any of them is 16\frac{1}{6}.

  4. 4.
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    If you see only the first digit, the last digit could have any of the 1010 possible values with equal probability. Similarly, if you see only the last digit, the first digit can have any of the 66 possible values with equal probability. Hence, the values of the two digits are independent. You can verify that the probability to see 00 is indeed 160\frac{1}{60} by multiplying the probabilities of each digit to be 0:

    16⋅110=160.\frac{1}{6}\cdot\frac{1}{10}=\frac{1}{60}.