
Tickets numbered 1 to 120 in a bag . What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
Answer
588.6k+ views
Hint: From the given data we get that the number of elements in the sample space is 120 and let A be the event of drawing a number of multiple 3 and B be a event of drawing a element of multiple 5 and we need to find the probability of these events and and their intersection and the required probability is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Complete step-by-step answer:
We are given that the tickets numbered from 1 to 120 are in a bag
Therefore our sample space $S = \left\{ {1,2,3,...,120} \right\}$
Therefore the number of elements in the sample space is 120
$ \Rightarrow n(S) = 120$
Let A be the event of drawing a number which is a multiple of 3
$ \Rightarrow A = \left\{
3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78, \\
81,84,87,90,93,96,99,102,105,108,111,114,117,120 \\
\right\}$ therefore n(A) = 40
Probability of drawing a number which is a multiple of 3
$ \Rightarrow P(A) = \dfrac{{40}}{{120}} = \dfrac{1}{3}$
Let B be the event of drawing a number which is a multiple of 5
$ \Rightarrow B = \left\{ {5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120} \right\}$therefore n(B) = 24
Probability of drawing a number which is a multiple of 5
$ \Rightarrow P(B) = \dfrac{{24}}{{120}} = \dfrac{1}{5}$
Now the elements common in both the sets are
$ \Rightarrow A \cap B = \left\{ {15,30,45,60,75,90,105,120} \right\}$
$ \Rightarrow n(A \cap B) = 8$
From this we get,
$ \Rightarrow P(A \cap B) = \dfrac{8}{{120}} = \dfrac{1}{{15}}$
The probability that the number drawn will be a multiple of 3 or 5 will be given by
$
\Rightarrow P(A \cup B) = P(A) + P(B) - P(A \cap B) \\
\Rightarrow P(A \cup B) = \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{{15}} \\
\Rightarrow P(A \cup B) = \dfrac{{5 + 3 - 1}}{{15}} = \dfrac{7}{{15}} \\
$
Therefore the probability that the number drawn will be a multiple of 3 or 5 is $\dfrac{7}{{15}}$
Note: 5 Basic Facts About Probability
1.A probability of 0 means that an event is impossible.
2.A probability of 1 means that an event is certain.
3.An event with a higher probability is more likely to occur.
4.Probabilities are always between 0 and 1.
5.The probabilities of our different outcomes must sum to 1.
Complete step-by-step answer:
We are given that the tickets numbered from 1 to 120 are in a bag
Therefore our sample space $S = \left\{ {1,2,3,...,120} \right\}$
Therefore the number of elements in the sample space is 120
$ \Rightarrow n(S) = 120$
Let A be the event of drawing a number which is a multiple of 3
$ \Rightarrow A = \left\{
3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78, \\
81,84,87,90,93,96,99,102,105,108,111,114,117,120 \\
\right\}$ therefore n(A) = 40
Probability of drawing a number which is a multiple of 3
$ \Rightarrow P(A) = \dfrac{{40}}{{120}} = \dfrac{1}{3}$
Let B be the event of drawing a number which is a multiple of 5
$ \Rightarrow B = \left\{ {5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120} \right\}$therefore n(B) = 24
Probability of drawing a number which is a multiple of 5
$ \Rightarrow P(B) = \dfrac{{24}}{{120}} = \dfrac{1}{5}$
Now the elements common in both the sets are
$ \Rightarrow A \cap B = \left\{ {15,30,45,60,75,90,105,120} \right\}$
$ \Rightarrow n(A \cap B) = 8$
From this we get,
$ \Rightarrow P(A \cap B) = \dfrac{8}{{120}} = \dfrac{1}{{15}}$
The probability that the number drawn will be a multiple of 3 or 5 will be given by
$
\Rightarrow P(A \cup B) = P(A) + P(B) - P(A \cap B) \\
\Rightarrow P(A \cup B) = \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{{15}} \\
\Rightarrow P(A \cup B) = \dfrac{{5 + 3 - 1}}{{15}} = \dfrac{7}{{15}} \\
$
Therefore the probability that the number drawn will be a multiple of 3 or 5 is $\dfrac{7}{{15}}$
Note: 5 Basic Facts About Probability
1.A probability of 0 means that an event is impossible.
2.A probability of 1 means that an event is certain.
3.An event with a higher probability is more likely to occur.
4.Probabilities are always between 0 and 1.
5.The probabilities of our different outcomes must sum to 1.
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