
Let A and B be two events such that $ P\left( \overline{A\cup B} \right)=\dfrac{1}{6} $ , $ P\left( A\cap B \right)=\dfrac{1}{4} $ and $ P\left( \overline{A} \right)=\dfrac{1}{4} $ , where $ \overline{A} $ = complementary event of A. Then A and B are:
A. Equally likely but not independent.
B. Equally likely and mutually exclusive.
C. Mutually exclusive and independent.
D. Independent but not equally likely.
Answer
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Hint:
1) If $ P\left( A \right)=P\left( B \right) $ , then the events are said to be equally likely.
2) If A and B are independent events, then $ P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right) $ .
3) If A and B are mutually exclusive events, then $ P\left( A\cap B \right)=0 $ .
4) For two events A and B, we have: $ P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) $ .
5) $ P\left( \overline{E} \right)=1-P\left( E \right) $ , where $ \overline{E} $ = complementary event of E.
Complete step by step solution:
Using the definition of complementary events:
$ P\left( A\cup B \right)=1-P\left( \overline{A\cup B} \right)=1-\dfrac{1}{6}=\dfrac{5}{6} $
$ P\left( A \right)=1-P\left( \overline{A} \right)=1-\dfrac{1}{4}=\dfrac{3}{4} $
Let us find out $ P\left( B \right) $ and examine the values of $ P\left( A\cap B \right) $ and $ P\left( A \right)\times P\left( B \right) $ to determine whether the events are equally likely and mutually exclusive or independent.
Using $ P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) $ , we get:
$ \dfrac{5}{6}=\dfrac{3}{4}+P\left( B \right)-\dfrac{1}{4} $
⇒ $ P\left( B \right)=\dfrac{5}{6}-\dfrac{3}{4}+\dfrac{1}{4}=\dfrac{1}{3} $
Since, $ P\left( A \right)\ne P\left( B \right) $ , the events are not equally likely.
Also, $ P\left( A \right)\times P\left( B \right)=\dfrac{3}{4}\times \dfrac{1}{3}=\dfrac{1}{4}=P\left( A\cap B \right) $ , so the events are independent.
The correct answer option is D. Independent but not equally likely.
Note:
1) A mutually exclusive event is defined as a situation where two events cannot occur at same time.
2) When a coin is tossed, there are two events possible, either it will be a Head or a Tail. Both the events here are mutually exclusive because they cannot happen simultaneously.
3) An independent event is where one event remains unaffected by the occurrence of the other event.
4) If we take two separate coins and flip them, then the occurrence of a Head or a Tail on both the coins are independent of each other, because a Head/Tail on one coin, does not affect the outcome of the other coin.
5) Mutually exclusive events are necessarily also dependent events because one's existence depends on the other's non-existence.
6) If at least one of the events has zero probability, then the two events can be mutually exclusive and independent simultaneously. However, if both events have non-zero probability, then they cannot be mutually exclusive and independent simultaneously.
1) If $ P\left( A \right)=P\left( B \right) $ , then the events are said to be equally likely.
2) If A and B are independent events, then $ P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right) $ .
3) If A and B are mutually exclusive events, then $ P\left( A\cap B \right)=0 $ .
4) For two events A and B, we have: $ P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) $ .
5) $ P\left( \overline{E} \right)=1-P\left( E \right) $ , where $ \overline{E} $ = complementary event of E.
Complete step by step solution:
Using the definition of complementary events:
$ P\left( A\cup B \right)=1-P\left( \overline{A\cup B} \right)=1-\dfrac{1}{6}=\dfrac{5}{6} $
$ P\left( A \right)=1-P\left( \overline{A} \right)=1-\dfrac{1}{4}=\dfrac{3}{4} $
Let us find out $ P\left( B \right) $ and examine the values of $ P\left( A\cap B \right) $ and $ P\left( A \right)\times P\left( B \right) $ to determine whether the events are equally likely and mutually exclusive or independent.
Using $ P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) $ , we get:
$ \dfrac{5}{6}=\dfrac{3}{4}+P\left( B \right)-\dfrac{1}{4} $
⇒ $ P\left( B \right)=\dfrac{5}{6}-\dfrac{3}{4}+\dfrac{1}{4}=\dfrac{1}{3} $
Since, $ P\left( A \right)\ne P\left( B \right) $ , the events are not equally likely.
Also, $ P\left( A \right)\times P\left( B \right)=\dfrac{3}{4}\times \dfrac{1}{3}=\dfrac{1}{4}=P\left( A\cap B \right) $ , so the events are independent.
The correct answer option is D. Independent but not equally likely.
Note:
1) A mutually exclusive event is defined as a situation where two events cannot occur at same time.
2) When a coin is tossed, there are two events possible, either it will be a Head or a Tail. Both the events here are mutually exclusive because they cannot happen simultaneously.
3) An independent event is where one event remains unaffected by the occurrence of the other event.
4) If we take two separate coins and flip them, then the occurrence of a Head or a Tail on both the coins are independent of each other, because a Head/Tail on one coin, does not affect the outcome of the other coin.
5) Mutually exclusive events are necessarily also dependent events because one's existence depends on the other's non-existence.
6) If at least one of the events has zero probability, then the two events can be mutually exclusive and independent simultaneously. However, if both events have non-zero probability, then they cannot be mutually exclusive and independent simultaneously.
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