$A$ and $B$ are two events where $P (A) = 0.25$ and $P (B) = 0.5$ . The probability of both happening together is $0.14$ . The probability of both $A$ and $B$ not happening is

0.39

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0.25

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0.11

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none of these

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Solution

The correct option is A $0.39$
$P (A \cap B) = 0.14 ∵ (A^{c} \cap B^{c}) = 1 - (A \cup B) T h e n P (A^{c} \cap B^{c}) = 1 - P (A \cup B) = 1 - {P (A) + P (B) - P (A \cap B)}$
Putting the given values
$P (A^{c} \cap B^{c}) = 1 - {0.25 + 0.5 - 0.14} = 0.39$

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Q. Two events

A

and

B

have the probabilities

0.25

and

0.5

respectively. The probability that both

A

and

B

occur simultaneously is 0.14. Then the probability of neither A nor B occurs is equal to:

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A and B are two events where P(A)=0.25 and P(B)=0.5. The probability of both happening together is 0.14. The probability of both A and B not happening is

The correct option is A 0.39P(A∩B)=0.14∵(Ac∩Bc)=1−(A∪B)ThenP(Ac∩Bc)=1−P(A∪B)=1−{P(A)+P(B)−P(A∩B)}Putting the given valuesP(Ac∩Bc)=1−{0.25+0.5−0.14}=0.39

$A$ and $B$ are two events where $P (A) = 0.25$ and $P (B) = 0.5$ . The probability of both happening together is $0.14$ . The probability of both $A$ and $B$ not happening is

The correct option is A $0.39$
$P (A \cap B) = 0.14 ∵ (A^{c} \cap B^{c}) = 1 - (A \cup B) T h e n P (A^{c} \cap B^{c}) = 1 - P (A \cup B) = 1 - {P (A) + P (B) - P (A \cap B)}$
Putting the given values
$P (A^{c} \cap B^{c}) = 1 - {0.25 + 0.5 - 0.14} = 0.39$