
There are three passengers on an airport shuttle bus that makes stops at four different hotels. The probability that all three passengers will be staying at different hotels, is-
(a) $ \dfrac{1}{{16}} $
(b) $ \dfrac{1}{4} $
(c) $ \dfrac{3}{8} $
(d) $ \dfrac{3}{4} $
Answer
530.1k+ views
Hint: Start by mentioning all the given conditions. Then evaluate each and every condition and according to that evaluate the values. After evaluating the values for each condition, put all the terms together and evaluate the final answer.
Complete step by step solution:
First we will start off by mentioning the total number of hotels and the number of passengers.
So, there are a total of $ 3 $ passengers and $ 4 $ hotels.
Now we will evaluate the number of ways in which the passengers can choose different hotels.
Now, here there are $ 4 $ hotels and $ 3 $ passengers, also here the decision of choosing the hotel lies totally on the passenger so that will be the deciding factor. So, the total number of ways of choosing $ 3 $ hotels among $ 4 $ hotels is $ {}^4{C_3} $ but the passengers choosing the hotel can also be rearranged. So, the total number of ways will be given by $ {}^4{C_3} \times 3! $ . If we evaluate its value, it comes out as $ 24 $ .
Now we will evaluate the total number of ways in which a passenger chooses a hotel.
So, a passenger can choose among $ 4 $ hotels so the number of ways will be $ 4 $ . Now, the next passenger can choose in
$ {}^3{C_2} \times {}^4{C_1} \times {}^3{C_1} $ and the number of ways for choosing the hotel for the third passenger will be $ {}^4{C_3} \times 3! $ .
Hence, the total number of ways will be
$ 4 + {}^3{C_2} \times {}^4{C_1} \times {}^3{C_1} + {}^4{C_3} \times 3! $ .
Now, if we calculate the value it comes out as $ 64 $
Now, we will evaluate the probability.
$
P = \dfrac{{24}}{{64}} \\
= \dfrac{3}{8} \;
$
Hence, the probability that all three passengers will be staying at different hotels is $ \dfrac{3}{8} $ which is option (C).
So, the correct answer is “Option C”.
Note: While mentioning conditions, make sure to mark all the important words so that you do not miss any important terms. Then next always evaluate all the conditions and then evaluate their values step by step to avoid any confusions. The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty.
Complete step by step solution:
First we will start off by mentioning the total number of hotels and the number of passengers.
So, there are a total of $ 3 $ passengers and $ 4 $ hotels.
Now we will evaluate the number of ways in which the passengers can choose different hotels.
Now, here there are $ 4 $ hotels and $ 3 $ passengers, also here the decision of choosing the hotel lies totally on the passenger so that will be the deciding factor. So, the total number of ways of choosing $ 3 $ hotels among $ 4 $ hotels is $ {}^4{C_3} $ but the passengers choosing the hotel can also be rearranged. So, the total number of ways will be given by $ {}^4{C_3} \times 3! $ . If we evaluate its value, it comes out as $ 24 $ .
Now we will evaluate the total number of ways in which a passenger chooses a hotel.
So, a passenger can choose among $ 4 $ hotels so the number of ways will be $ 4 $ . Now, the next passenger can choose in
$ {}^3{C_2} \times {}^4{C_1} \times {}^3{C_1} $ and the number of ways for choosing the hotel for the third passenger will be $ {}^4{C_3} \times 3! $ .
Hence, the total number of ways will be
$ 4 + {}^3{C_2} \times {}^4{C_1} \times {}^3{C_1} + {}^4{C_3} \times 3! $ .
Now, if we calculate the value it comes out as $ 64 $
Now, we will evaluate the probability.
$
P = \dfrac{{24}}{{64}} \\
= \dfrac{3}{8} \;
$
Hence, the probability that all three passengers will be staying at different hotels is $ \dfrac{3}{8} $ which is option (C).
So, the correct answer is “Option C”.
Note: While mentioning conditions, make sure to mark all the important words so that you do not miss any important terms. Then next always evaluate all the conditions and then evaluate their values step by step to avoid any confusions. The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty.
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