Choosing balls
Task number: 3535
A bag contains 10 glass balls and 20 clay balls. We choose 7 balls at random. What is the probability that exactly 3 glass balls will be chosen if:
Variant
We do not return balls to the bag.
Hint
To design a probability space correctly, it's convenient to assume that the balls can be distinguished in some additional way, e.g. by color.
Solution
The set of elementary events consists of all possible choices of seven balls from thirty. This gives us \(|\Omega|=\binom{30}{7}\). We consider this to be a uniform probability distribution.
The event in question corresponds to the choices where we first choose three elements from a ten-element set of glass balls and then choose four clay balls from a set of twenty.
\(P(A)=\frac{|A|}{|\Omega|}=\frac{\binom{10}{3}\binom{20}{4}}{\binom{30}{7}}=\frac{120\,\cdot\, 4{,}845}{20{,}350,800}\doteq 29\%\)
Answer
The probability of the given event is \(P(A)=\frac{\binom{10}{3}\binom{20}{4}}{\binom{30}{7}}\doteq 29\%\).
Variant
We choose balls one by one and return each one to the bag immediately.
Solution
Now the probability space is formed by all mappings from a seven-element set of trials to a thirty-element set of balls, i.e. \(|\Omega|=30^7\).
The desired event consists of those mappings whose range of values contains three glass balls and four clay balls. The number of these is \(|B|=\binom{7}{3} 10^3 {\cdot} 20^4\). The first factor corresponds to the choice of draws in which we choose a glass ball, the second to the concrete choice of glass balls and the third to the choice of clay balls.
\(P(B)=\frac{\binom{7}{3}\cdot 10^3{\cdot} 20^4}{30^7}=\frac{560}{2{,}187}\doteq 26\%\)
\medskip Alternatively we can make the following argument. The probability of choosing one glass ball is \(\frac{10}{30}\). Choosing three glass balls has probability \(\left(\frac{1}{3}\right)^3\). To choose exactly three glass balls, we must also choose clay balls four times, and each such choice has probability \(\frac{2}{3}\). There are a total of \(\binom{7}{3}\) possibilities determining which three of the seven draws a glass ball is to be chosen in. The individual draws are independent. So we have
\(P(B)=\binom{7}{3}\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^4\).
Answer
The probability of the given event is \(P(B)=\binom{7}{3}\frac{2^4}{3^7}=\frac{560}{2{,}187}\doteq 26\%\).
