Question

A man takes a step forward with probability 0.4 and backward with probability 0.6. Find the probability that at the end of 5 steps, he is one step away from the starting point.

OR

Suppose a girl throws a die. If she gets a 1 or 2, she tosses a coin three times and notes the number of "tails". If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a 'head' or 'tail' is obtained. If she obtained exactly one "tail", what is the probability that she threw 3, 4, 5 or 6 with the die?

Answer 1

Given probability that the man takes a step forward = 0.4 = p so, q = 1 - p = 0.6.

Let$E_1$ be the event that out of 5 steps the man takes exactly 3 steps are forward and 2 steps backward.

Let $E_2$ be the event that out of 5 steps tha man takes exactly 2 steps are forward and 3 steps backward.

Also let E be the event that at the end of 5 steps, the man is one step away from the starting point.

$\therefore P\left(E\right)=P\left(E_1\right)+P\left(E_2\right)+P\left(3\right)+P\left(2\right){=}^5{C}_3\left(0.4{)}^3\right(0.6{)}^2{}^{}{+}^5{C}_2\left(0.4{)}^2\right(0.6{)}^3[\therefore P(r\left){=}^n{C}_r\right(p{)}^r\left(q{)}^{n-1}\right]$

$\Rightarrow =10\left(0.4{)}^3\right(0.6{)}^2+10\left(0.4{)}^2\right(0.6{)}^3=10\times \left(0.4{)}^2\right(0.6{)}^2[0.4+0.6]=10\times (0.24{)}^2=\mathrm{0.576.}$

OR

Let A : getting 1 or 2, B : getting 3, 4, 5 or 6, E : getting exactly one tail.

So, $P\left(A\right)=\frac{2}{6},P\left(B\right)=\frac{4}{6},P\left(\frac{E}{A}\right){=}^3{C}_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{\frac{3}{-1}}=\frac{3}{8}$ (using binomial distribution), $P\left(\frac{E}{B}\right)=\frac{1}{2}.$

By Bayes' theorem, $P\left(\frac{B}{E}\right)=\frac{P\left(\frac{E}{B}\right)P\left(B\right)}{P\left(\frac{E}{A}\right)P\left(A\right)+P\left(\frac{E}{B}\right)P\left(B\right)}$

$\therefore P\left(\frac{B}{E}\right)=\frac{\frac{1}{2}\times \frac{4}{6}}{\frac{3}{8}\times \frac{2}{6}+\frac{1}{2}\times \frac{4}{6}}=\frac{8}{3+8}=\frac{8}{11}.$