a bag contains 5 red and 10 white balls. a ball is drawn at random and is replaced after noting the colour. if 3 balls are drawn at random what is the probability that it is red.
A bag contains 5 red and 10 white balls, totaling 15 balls. Since each drawn ball is replaced after noting its color, the draws are independent, and the probability of drawing a red ball remains constant at each step.
Probability Calculation
The probability of drawing a red ball on a single draw is 515=13\frac{5}{15}=\frac{1}{3}155=31. For 3 independent draws, the probability of getting exactly one red (most common interpretation, as "it is red" often implies at least one in probability problems) requires binomial probability: (31)(13)1(23)2=3×13×49=49\binom{3}{1}\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^2=3\times \frac{1}{3}\times \frac{4}{9}=\frac{4}{9}(13)(31)1(32)2=3×31×94=94.
If "that it is red" means all 3 red , it's simply (13)3=127\left(\frac{1}{3}\right)^3=\frac{1}{27}(31)3=271.
Step-by-Step Breakdown
- Total balls: 15 (5 red + 10 white).
- P(red) = 515=13\frac{5}{15}=\frac{1}{3}155=31 each time (replacement keeps odds fixed).
- For independent events, multiply probabilities across draws.
Binomial Scenarios
Number of Reds| Probability Formula| Exact Value
---|---|---
Exactly 1| (31)p1(1−p)2\binom{3}{1}p^1(1-p)^2(13)p1(1−p)2| 49\frac{4}{9}942
Exactly 2| (32)p2(1−p)1\binom{3}{2}p^2(1-p)^1(23)p2(1−p)1| 29\frac{2}{9}92
All 3| p3p^3p3| 127\frac{1}{27}271
At least 1| 1−(1−p)31-(1-p)^31−(1−p)3| 2627\frac{26}{27}2726
TL;DR: Most likely exactly one red : 49\frac{4}{9}94 (~44.4%).
