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

Hint: 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 \[P\left( A \right) \cdot P\left( B \right)\].