probability

Question

There are 3 fair coins and 1 false coin with tails on both sides. A coin is chosen at random and tossed 4 times. if tail occures in all 4 times, then the probability that the false coin has been chosen for tossing is ?

Answer 1

$$\begin{align*} & \Rightarrow \text{Required probability} = \frac{\text{Probability along required path}}{\sum_{\color{red}{\bf \text{all path}}} \left ( \text{Probability along a path} \right )} \\ \\ & \Rightarrow \text{Required probability} = \frac{\frac{1}{4}.1}{\frac{1}{4}.1 + 3.\frac{1}{4}.\frac{1}{16}} \\ \\ & \Rightarrow \text{Required probability} = \frac{\frac{1}{4}}{\frac{1}{4}.\left [ 1+\frac{3}{16} \right ]} \\ \\ & \Rightarrow \text{Required probability} = {\large \color{red}{\frac{16}{19}}} \\ \end{align*}$$

Answer 2

B: Bised coin

F: Fair coin

Four tails: T

We should find P( B | T )

Bayes theorem works fine here,

Probability of choosing coin with no tosses

P(B)=1/4

p(F)=3/4

Probability of of 4 tails in each case,

P( T | B )=1 (always tails)

P( T | F)=1/16

P(B | T)= P( T | B) P(B) / P(B)P(T | B ) + P(F) P(T | F)

=16/19