First slide

Theorems of probability

Question

bag contains 20 coins. If the probability that the bag contains exactly 4 biased coin is 1/3 and that of exactly 5 biased coin is 2/3, then the probability that all the biased coin are sorted out from the bag in exactly 10 draws is

Moderate

A

$\frac{5}{10} \frac{^{16} C_{6}}{^{20} C_{9}} + \frac{1}{11} \frac{^{15} C_{5}}{^{20} C_{9}}$
B

$\frac{2}{33} [\frac{^{16} C_{6} + 5^{15} C_{5}}{^{20} C_{9}}]$
C

$\frac{5}{33} \frac{^{16} C_{7}}{^{20} C_{9}} + \frac{1}{11} \frac{^{15} C_{6}}{^{20} C_{9}}$
D

none of these

Solution

P(4 biased coin) $= \frac{1}{3}$
P(5 biased coin) $= \frac{1}{4}$
Hence, the required probability is
$\begin{array}{l} = \frac{1}{3} \frac{^{4} {C_{3}}^{16} C_{6}}{^{20} C_{9}} + \frac{2}{3} \frac{^{5} {C_{4}}^{15} C_{5}}{^{20} C_{9}} \frac{1}{^{11} C_{1}} \\ = \frac{2}{33} [\frac{^{16} C_{6} + 5^{15} C_{5}}{^{20} C_{9}}] \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App