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Solve the following.
An urn contains 25 balls of which 10 balls are red and the remaining green. A ball is drawn at…
An urn contains 25 balls of which 10 balls are red and the remaining green. A ball is drawn at random from the urn, the colour is noted and the ball is replaced. If 6 balls are drawn in this way, find the probability that:
(i)[1.6666666666666667]
All the balls are red.
(ii)[1.6666666666666667]
Not more than 2 balls are green.
(iii)[1.6666666666666667]
Number of red balls and green balls are equal.
Answer
Answer (i)
AI$P(\text{red}) = p = \frac{10}{25} = \frac{2}{5}$, $q = \frac{3}{5}$, $n = 6$. $P(X = r) = {}^6C_r\,p^r q^{6-r}$.
$P(\text{all red}) = {}^6C_6\left(\frac{2}{5}\right)^6$
Hence the probability is $\frac{64}{15625}$.
Final answer: $\frac{64}{15625}$
Answer (ii)
AINot more than 2 green means at least 4 red. Let $Y$ be the number of green balls, $P(\text{green}) = \frac{3}{5}$.
$P(Y \le 2) = {}^6C_0\left(\frac{2}{5}\right)^6 + {}^6C_1\left(\frac{3}{5}\right)\left(\frac{2}{5}\right)^5 + {}^6C_2\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^4$
$= \frac{64 + 576 + 2160}{15625} = \frac{2800}{15625}$
Hence the probability is $\frac{112}{625}$.
Final answer: $\frac{112}{625}$
Answer (iii)
AIEqual numbers means 3 red and 3 green.
$P = {}^6C_3\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^3 = 20\times\frac{8\times 27}{15625} = \frac{4320}{15625}$
Hence the probability is $\frac{864}{3125}$.
Final answer: $\frac{864}{3125}$
From ISC 2017 Mathematics Paper 1, question 12(b).