Question

1. An urn contains 13 balls numbering from 1 to 13. Find the probability that a ball selected at random is a ball with a number that is a multiple of 3 or 4.
2. The probability that a contractor will get a plumbing contract is \[\dfrac{2}{3}\] and an electric contract is \[\dfrac{4}{9}\]. If the probability of getting at least one contract is \[\dfrac{4}{5}\], find the probability that he will get both the contracts.

Answer 1

Hint: In this type of question we have to use the concept of probability. We know that the probability of an event is given by, \[\text{Probability = }\dfrac{\text{No}\text{. of favourable outcomes}}{\text{No}\text{. of all possible outcomes}}\].
In the first part we have given 13 balls numbered from 1 to 13 , so we find out the numbers which are divisible by 3 or 4 or both and then by using the above formula we can obtain the required result.
In second part we have already given the probabilities for two different events along with the probability of at least one event and we have to find out the probability for both events, so we have to use the general probability addition rule for the union of two events which states that \[P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)\].

Complete step by step answer:
1. Now, we have to find the probability that a randomly selected ball from an urn consisting of 13 balls numbered from 1 to 13 is with a number that is multiple of 3 or 4.
Let us first list out all the numbers from 1 to 13 which are either multiple of 3 or 4,
\[\Rightarrow A=\left\{ 3,4,6,8,9,12 \right\}\]
Hence, we get 6 balls with numbers that are multiple of 3 or 4.
\[\Rightarrow \text{No}\text{. of favourable outcomes}=6\]
Also we know that the urn consists of 13 balls numbering from 1 to 13
\[\Rightarrow \text{No}\text{. of all possible outcomes}=13\]
As we know that, the probability is given by
\[\Rightarrow \text{Probability = }\dfrac{\text{No}\text{. of favourable outcomes}}{\text{No}\text{. of all possible outcomes}}\]
Thus, the probability of getting a ball with a number that is multiple of 3 or 4 is given by
\[\Rightarrow \text{Probability = }\dfrac{6}{13}\].
2. Now, here the probability that a contractor will get a plumbing contract is \[\dfrac{2}{3}\], an electric contract is \[\dfrac{4}{9}\] and the probability of getting at least one contract is \[\dfrac{4}{5}\], and we have to find the probability that he will get both the contracts.
Let us suppose that, the event A will represent the contractor will get a plumbing contract, B will represent the contractor will get electric contract, then by given we can write
\[\begin{align}
  & \Rightarrow P\left( A \right)=\dfrac{2}{3} \\
 & \Rightarrow P\left( B \right)=\dfrac{4}{9} \\
\end{align}\]
Thus \[A\bigcup B\] will represent that the contractor will get at least one contract
\[\Rightarrow P\left( A\bigcup B \right)=\dfrac{4}{5}\]
We know that by general probability addition rule for the union of two events states that
\[\Rightarrow P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)\]
As we have to find the probability that the contractor will get both the contracts, we have to find \[P\left( A\bigcap B \right)\]. Hence by rewriting the above rule we get,
\[\Rightarrow P\left( A\bigcap B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcup B \right)\]
By substituting the given probabilities we can write
\[\begin{align}
  & \Rightarrow P\left( A\bigcap B \right)=\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{4}{5} \\
 & \Rightarrow P\left( A\bigcap B \right)=\dfrac{2\times 15}{3\times 15}+\dfrac{4\times 5}{9\times 5}-\dfrac{4\times 9}{5\times 9} \\
 & \Rightarrow P\left( A\bigcap B \right)=\dfrac{30+20-36}{45} \\
 & \Rightarrow P\left( A\bigcap B \right)=\dfrac{14}{45} \\
\end{align}\]
Hence, the probability that the contractor will get both the contracts is \[\dfrac{14}{45}\].

Note: In the first part of this question students have to consider the numbers which are multiples of 3 and 4 and both. Students can also solve the first part as follows:
Let us suppose A be the event that the number is divisible by 3
\[\Rightarrow A=\left\{ 3,6,9,12 \right\}\]
Let us suppose B be the event that the number is divisible by 4
\[\Rightarrow B=\left\{ 4,8,12 \right\}\]
Hence, we can say that \[A\bigcup B\] will represent the set of numbers which are divisible by 3 or 4
\[\Rightarrow A\bigcup B=\left\{ 3,4,6,8,9,12 \right\}\]
\[\Rightarrow n\left( A\bigcup B \right)=6\]
In other words we can say that
\[\Rightarrow \text{No}\text{. of favourable outcomes}=6\]
Also we know that the urn consists of 13 balls numbering from 1 to 13
\[\Rightarrow \text{No}\text{. of all possible outcomes}=13\]
As we know that, the probability is given by
\[\Rightarrow \text{Probability = }\dfrac{\text{No}\text{. of favourable outcomes}}{\text{No}\text{. of all possible outcomes}}\]
Thus, the probability of getting a ball with a number that is multiple of 3 or 4 is given by
\[\Rightarrow \text{Probability = }\dfrac{6}{13}\].
In the second part of the question, students have to remember the general probability addition rule for the union of two events. Also students have to note that the phrase ‘at least one event’ will represent the union of two events while the phrase ‘both events’ will represent intersection of two events.