SOLUTION: If an urn has 11 white balls and 9 red balls, what is the probability that 5 randomly selected balls contain at least 3 white balls? (Enter answer as a decimal with at least 4 corr
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Question 1209797: If an urn has 11 white balls and 9 red balls, what is the probability that 5 randomly selected balls contain at least 3 white balls? (Enter answer as a decimal with at least 4 correct decimal places) Found 2 solutions by CPhill, greenestamps:Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let $W$ be the number of white balls and $R$ be the number of red balls in the urn.
We have $W = 11$ and $R = 9$, so the total number of balls is $N = W + R = 11 + 9 = 20$.
We want to find the probability of selecting 5 balls with at least 3 white balls. This means we can have 3, 4, or 5 white balls.
The total number of ways to select 5 balls from 20 is given by $\binom{20}{5}$.
$$ \binom{20}{5} = \frac{20!}{5!15!} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = 15504 $$
Now, let's calculate the number of ways to select:
\begin{itemize}
\item 3 white balls and 2 red balls: $\binom{11}{3} \binom{9}{2} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} \times \frac{9 \times 8}{2 \times 1} = 165 \times 36 = 5940$
\item 4 white balls and 1 red ball: $\binom{11}{4} \binom{9}{1} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} \times 9 = 330 \times 9 = 2970$
\item 5 white balls and 0 red balls: $\binom{11}{5} \binom{9}{0} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} \times 1 = 462 \times 1 = 462$
\end{itemize}
The total number of ways to select at least 3 white balls is:
$$ 5940 + 2970 + 462 = 9372 $$
The probability of selecting at least 3 white balls is:
$$ \frac{9372}{15504} = \frac{2343}{3876} \approx 0.6044889061 $$
Therefore, the probability is approximately 0.6045.
Final Answer: The final answer is $\boxed{0.6045}$
