Question

A coin is tossed. If head appears a fair die is thrown three times otherwise a biased die with probability of obtaining an even number twice as that of an odd number is thrown three times. If $(n_1,n_2,n_3)$ is an outcome, $(1\le n_1,n_2,n_3\le 6)$ and is found to satisfy the equation $i^{n_1}+i^{n_2}+i^{n_3}=1,$ then the probability that the fair die was thrown is (where $i=\sqrt{-1}$)

• A. $\frac{1}{12}$
• B. $\frac{1}{3}$
• C. $\frac{27}{59}$
• D. None of these

Answer 1

The correct option is C $\frac{27}{59}$

Let $E$ be an event $F$ represents tossing of a fair die and $B$ represents tossing of a biased die.
$P\left(F|\right(n_1,n_2,n_3\left)E\right)=\frac{P\left(\right(n_1,n_2,n_3\left)E|F\right)\times P\left(F\right)}{P\left(\right(n_1,n_2,n_3\left)E|F\right)\times P\left(F\right)+P\left(\right(n_1,n_2,n_3\left)E|B\right)\times P\left(B\right)}$
$P\left(F\right)=P\left(B\right)=\frac{1}{2}$
$P\left(1\right)=P\left(2\right)\cdots =P\left(6\right)=\frac{1}{6}$ (for a fair dice.)
$2\times P\left(\right\{1,3,5\}=P(\{2,4,6\})$ (for biased dice)
$=1-P\left(\right\{1,3,5\left\}\right)$
$P\left(\right\{1,3,5\}=\frac{1}{3}$
$P\left(\right\{2,4,6\left\}\right)=\frac{2}{3}$
Assume that $1,3,5$ are equally likely for biased dice and same for $2,4,6$
$P\left(1\right)=P\left(3\right)=P\left(5\right)=\frac{1}{9}$
$P\left(2\right)=P\left(4\right)=P\left(6\right)=\frac{2}{9}{i}^{n_1}+i^{n_2}+i^{n_3}=1$
We can have 2 possible cases
$(1,1,-1)$ or $i,-i,1$
Permutations of $n_1,n_2,n_3$ will give
$n_1=4,n_2=4,n_3=(2,6)$
$n_1,n_2,n_3$$\to (4,4,2)\text{or}(4,4,6)$
Each of these cases can be permutated in number of ways
$=\frac{3!}{2!}=3$
For fair dice :
$P(n_1=4,n_2=4,n_3=(2,6\left)|F\right)=6\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$
For biased die
$(4,4,2)(4,4,6)$
$\frac{2}{9}\times \frac{2}{9}\times \frac{2}{9}\frac{2}{9}\times \frac{2}{9}\times \frac{2}{9}$
$i^{n_1}+i^{n_2}+i^{n_3}=14$
$(i,-i,1)$

$n_1{n}_2{n}_3$
$(1,5)34$
$(1,3,4)(5,3,4)$
6 ways 6 ways
Fair$12(\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6})=\frac{2}{36}$

B$\to 12(\frac{1}{9}\times \frac{1}{9}\times \frac{2}{9})=\frac{8}{243}$
$P\left(\right(n_1,n_2,n_3\left)E/F\right)=\frac{3}{36}=\frac{1}{12}$
$P\left(\right(n_1,n_2,n_3\left)E/B\right)=\frac{3}{36}=\frac{16}{243}+\frac{8}{243}=\frac{24}{243}$
$P\left(F/\right(n_1,n_2,n_3\left)E\right)=\frac{\frac{1}{12}\times \frac{1}{2}}{\frac{1}{12}\times \frac{1}{2}+\frac{24}{243}\times \frac{1}{2}}$
$=\frac{\frac{1}{4}}{\frac{27+32}{27\times 4}}=\frac{27}{59}$