Mar 30

Calculating the probability of an event by its complement

By Tutor GuyNo Comments

 

You will often be asked to calculate the probability of a compound event; that is, an event that contains two or more simple outcomes. For example, if you flip five coins, what is the probability that you get either four or five heads? To solve this problem, you calculate the probability of getting exactly four heads and the probability of getting exactly five heads, then add the numbers together:

P(x=4 \text{ or } x=5)=P(x=4)+P(x=5)=

\displaystyle \binom{5}{4} \! \left ( \dfrac{1}{2} \right )^4 \!\! \left ( \dfrac{1}{2} \right)^1+ \binom{5}{5} \! \left (\dfrac{1}{2} \right )^5 \!\! \left ( \dfrac{1}{2} \right )^0= \dfrac{5}{32}+ \dfrac{1}{32}= \dfrac{6}{32}= \dfrac{3}{16}

But what if a compound event contains a lot of simple events? For example, if you roll ten dice, what is the probability you get at least two sixes? To solve this the way we did the previous example, we would need to find the probability of getting exactly two sixes, the probability of getting exactly three sixes, and so on, up to the probability of getting exactly 10 sixes. Although the calculations are not difficult, it is very tedious to find nine different probabilities in order to add them all together. For this problem, it is much simpler to calculate the complement of the given event. The complement of “at least two sixes” is “at most one six”. So calculate the probability of getting at most one six when you roll ten dice:

P(x=0 \text{ or } x=1)=P(x=0)+P(x=1)=

\displaystyle \binom{10}{0} \! \left ( \frac{1}{6} \right )^0 \!\! \left ( \frac{5}{6} \right )^{10}+ \binom{10}{1} \! \left ( \frac{1}{6} \right )^1 \!\! \left ( \frac{5}{6} \right )^9=0.1615+0.3230=0.4845

Because the probability of an event A and its complement (not A) add up to 1, the probability of getting at least two sixes is then 1-0.4845=0.5155.

Calculating a compound probability by finding the probability of the complement instead is the basis of many problems in statistics. It’s up to you to determine when this is the best approach to solving a problem.

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