Question

Shalu brought 2 cages of birds: cage-I contains 5 parrots and 1 owl and cage-II contains 6 parrots. One day, Shalu forgot to lock both the cages and 2 birds flew from cage-I to cage-II. Then 2 birds flew back from cage-II to cage-I. Birds moved one after the other and not simultaneously. Assume that all birds have equal chance of flying, the probability that owl is still in cage 1 is:
A.$\dfrac{1}{6}$
B.$\dfrac{1}{3}$
C.$\dfrac{2}{3}$
D.$\dfrac{3}{4}$

Answer 1

We will find the probability of the owl not flying from cage-I and the probability of the owl flying to cage-II and then returning to cage-I. We will add these 2 probabilities to find the answer.

Formulas used:
We will use the formula of the probability (P) is given by ${\rm{P}} = \dfrac{{\rm{F}}}{{\rm{T}}}$ where F is the number of favourable outcomes and T is the number of total outcomes.

Complete step-by-step answer:
The owl will remain in cage-I if (A) either it does not fly away from cage-I or (B)if it flies to cage-II and flies backs to cage-I.
Event A:
First, we will find the probability that the owl doesn’t fly from the 1st cage; that is, only the parrots fly from cage-I.
There are 5 parrots and 6 birds in total in cage-I.
Event C:
The probability that the 1st bird that flies is a parrot is $\dfrac{5}{6}$.
Now, there are 4 parrots and 5 birds left in cage-I.
Event D:
The probability that the 2nd bird to fly will be a parrot is $\dfrac{4}{5}$.
We will use formula $P\left( A \right) \cdot P\left( B \right)$ to calculate the probability that the owl doesn’t fly from the cage:
$P\left( C \right) \cdot P\left( D \right) = \dfrac{5}{6} \times \dfrac{4}{5} = \dfrac{4}{6}$
Dividing both numerator and denominator by 2, we get
$\Rightarrow P\left( C \right) \cdot P\left( D \right) = \dfrac{2}{3}$
Event B:
Now, we will find the probability that the owl flies to cage-II and then flies back to cage-I.
The probability that the owl flies first and a parrot next is $\dfrac{1}{6} \times \dfrac{5}{5}$ . The probability that a parrot flies first and the owl next is $\dfrac{5}{6} \times \dfrac{1}{5}$.
Event E:
The probability that the owl flies from cage-I to cage-II $= \left( {\dfrac{1}{6} \times \dfrac{5}{5}} \right) + \left( {\dfrac{5}{6} \times \dfrac{1}{5}} \right) = \dfrac{1}{6} + \dfrac{1}{6}$
Simplifying the expression, we get
$\Rightarrow$The probability that the owl flies from cage-I to cage-II $= \dfrac{1}{3}$
Now we will use formula $P\left( A \right) \cdot P\left( B \right)$ to get the probabilities.
The probability that the owl flies back first and a parrot next is $\dfrac{1}{8} \times \dfrac{7}{7}$ .
The probability that a parrot flies back first and the owl next is $\dfrac{7}{8} \times \dfrac{1}{7}$.
Event F:
We will use formula $P\left( A \right) + P\left( B \right)$ to find the probability that the owl flies back from cage-II to cage-I:
The probability that the owl flies back from cage-II to cage-I $= \left( {\dfrac{1}{8} \times \dfrac{7}{7}} \right) + \left( {\dfrac{7}{8} \times \dfrac{1}{7}} \right) = \dfrac{1}{8} + \dfrac{1}{8}$
Adding the terms, we get
$\Rightarrow$ The probability that the owl flies back from cage-II to cage-I $= \dfrac{1}{4}$
The probability that the owl flies to cage-II and then flies back to cage-I is:
$P\left( E \right) \cdot P\left( F \right) = \dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{{12}}$
The probability that the owl does not fly away from cage-I or flies to cage-II and then flies backs to cage-I is:
$P\left( A \right) + P\left( B \right) = \dfrac{2}{3} + \dfrac{1}{{12}}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow P\left( A \right) + P\left( B \right) = \dfrac{{8 + 1}}{{12}}\\\ \Rightarrow P\left( A \right) + P\left( B \right) = \dfrac{9}{{12}}\end{array}$
Dividing both numerator and denominator by 3, we get
$\Rightarrow P\left( A \right) + P\left( B \right) = \dfrac{3}{4}$
$\therefore$ Option D is the correct option.

Note: Probability is a certainty of occurrence of an event. It tells us whether the event will occur or not. If the probability of event $A$ happening is $P\left( A \right)$ and the probability of event $B$ happening is $P\left( B \right)$, then:
(i) The probability that either $A$ or $B$ will happen is $P\left( A \right) + P\left( B \right)$.
(ii) The probability that $A$ and $B$ will happen simultaneously is $P\left( A \right) \cdot P\left( B \right)$.