
The probability that a new car will get an award for its design is \[0.25\], the probability that it will get an award for efficient use of fuel is \[0.35\] and the probability that it will get both the awards is \[0.15\]. Find the probability that
A) it will get at least one of the two awards.
B) it will get only one of the awards.
Answer
507.9k+ views
Hint:
For (i), we will use the property of probability, \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\], where \[P\left( A \right)\] represents the probability that a new car will get an award for its design, \[P\left( B \right)\] represents the probability that a new car will get an award for efficient use of fuel and \[P\left( {A \cap B} \right)\] represents the probability that a new car will get both the awards to find the required value. Then for (ii), we will use the property,\[P\left( {A' \cap B} \right) + P\left( {A \cap B'} \right)\]and \[P\left( {A' \cap B} \right) = P\left( A \right) - P\left( {A \cap B} \right)\], where \[A'\] be not for the design of the car and \[B'\] be not for the efficient use of fuel in the car to find the required value.
Complete step by step solution:
Let us assume that \[A\] be for the design of the car and \[B\] be for the efficient use of fuel in the car.
Let us also assume that \[P\left( A \right)\] represents the probability that a new car will get an award for its design.
Since we are given that the probability that a new car will get an award for its design is \[0.25\], then \[P\left( A \right) = 0.25\].
Let us assume that \[P\left( B \right)\] represents the probability that a new car will get an award for efficient use of fuel.
Since we are given that the probability that a new car will get an award for efficient use of fuel is \[0.35\], then \[P\left( B \right) = 0.35\].
Let us assume that \[P\left( {A \cap B} \right)\] represents the probability that a new car will get both the awards.
Since we are given that the probability that a new car will get an award for both is \[0.15\], then \[P\left( {A \cap B} \right) = 0.15\].
We will now find the probability of getting at least one of the awards, that is, \[P\left( {A \cup B} \right)\].
We will now use the property of probability, that is, \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\] to find the probability that the car will get at least one of the two awards.
Substituting these values of \[P\left( A \right)\], \[P\left( B \right)\] and \[P\left( {A \cap B} \right)\] in the above property of probability, we get
\[
P\left( {A \cup B} \right) = 0.25 + 0.35 - 0.15 \\
= 0.60 - 0.15 \\
= 0.45 \\
\]
Thus, the probability of at least one of the two awards is \[0.45\].
Let us assume that \[A'\] be not for the design of the car and \[B'\] be not for the efficient use of fuel in the car.
We will now find the probability of getting only one of the awards.
\[P\left( {A' \cap B} \right) + P\left( {A \cap B'} \right)\]
We will now use the property of probability, that is,
\[P\left( {A' \cap B} \right) = P\left( A \right) - P\left( {A \cap B} \right)\] to find the probability that the car gets only one of the awards.
Simplifying the above equation using the above property of probability, we get
\[ \Rightarrow P\left( A \right) - P\left( {A \cap B} \right) + P\left( B \right) - P\left( {A \cap B} \right)\]
Substituting these values of \[P\left( A \right)\], \[P\left( B \right)\] and \[P\left( {A \cap B} \right)\] in the above property of probability, we get
\[
\Rightarrow \left( {0.25 - 0.15} \right) + \left( {0.35 - 0.15} \right) \\
\Rightarrow 0.10 + 0.20 \\
\Rightarrow 0.30 \\
\]
Thus, the probability of getting only one of the awards is \[0.30\].
Note:
In solving these types of questions, you should be familiar with the formula to find the probability of any event happening and not happening. Some students get confused while applying formulae. In this question, one can find the probability for exactly one award instead of at least one of the awards and then conclude the wrong answer. It means that we have to find the probability that the minimum one award should be received.
For (i), we will use the property of probability, \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\], where \[P\left( A \right)\] represents the probability that a new car will get an award for its design, \[P\left( B \right)\] represents the probability that a new car will get an award for efficient use of fuel and \[P\left( {A \cap B} \right)\] represents the probability that a new car will get both the awards to find the required value. Then for (ii), we will use the property,\[P\left( {A' \cap B} \right) + P\left( {A \cap B'} \right)\]and \[P\left( {A' \cap B} \right) = P\left( A \right) - P\left( {A \cap B} \right)\], where \[A'\] be not for the design of the car and \[B'\] be not for the efficient use of fuel in the car to find the required value.
Complete step by step solution:
Let us assume that \[A\] be for the design of the car and \[B\] be for the efficient use of fuel in the car.
Let us also assume that \[P\left( A \right)\] represents the probability that a new car will get an award for its design.
Since we are given that the probability that a new car will get an award for its design is \[0.25\], then \[P\left( A \right) = 0.25\].
Let us assume that \[P\left( B \right)\] represents the probability that a new car will get an award for efficient use of fuel.
Since we are given that the probability that a new car will get an award for efficient use of fuel is \[0.35\], then \[P\left( B \right) = 0.35\].
Let us assume that \[P\left( {A \cap B} \right)\] represents the probability that a new car will get both the awards.
Since we are given that the probability that a new car will get an award for both is \[0.15\], then \[P\left( {A \cap B} \right) = 0.15\].
We will now find the probability of getting at least one of the awards, that is, \[P\left( {A \cup B} \right)\].
We will now use the property of probability, that is, \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\] to find the probability that the car will get at least one of the two awards.
Substituting these values of \[P\left( A \right)\], \[P\left( B \right)\] and \[P\left( {A \cap B} \right)\] in the above property of probability, we get
\[
P\left( {A \cup B} \right) = 0.25 + 0.35 - 0.15 \\
= 0.60 - 0.15 \\
= 0.45 \\
\]
Thus, the probability of at least one of the two awards is \[0.45\].
Let us assume that \[A'\] be not for the design of the car and \[B'\] be not for the efficient use of fuel in the car.
We will now find the probability of getting only one of the awards.
\[P\left( {A' \cap B} \right) + P\left( {A \cap B'} \right)\]
We will now use the property of probability, that is,
\[P\left( {A' \cap B} \right) = P\left( A \right) - P\left( {A \cap B} \right)\] to find the probability that the car gets only one of the awards.
Simplifying the above equation using the above property of probability, we get
\[ \Rightarrow P\left( A \right) - P\left( {A \cap B} \right) + P\left( B \right) - P\left( {A \cap B} \right)\]
Substituting these values of \[P\left( A \right)\], \[P\left( B \right)\] and \[P\left( {A \cap B} \right)\] in the above property of probability, we get
\[
\Rightarrow \left( {0.25 - 0.15} \right) + \left( {0.35 - 0.15} \right) \\
\Rightarrow 0.10 + 0.20 \\
\Rightarrow 0.30 \\
\]
Thus, the probability of getting only one of the awards is \[0.30\].
Note:
In solving these types of questions, you should be familiar with the formula to find the probability of any event happening and not happening. Some students get confused while applying formulae. In this question, one can find the probability for exactly one award instead of at least one of the awards and then conclude the wrong answer. It means that we have to find the probability that the minimum one award should be received.
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