Let A- B- C be three events in a probability space- Suppose that PA-0-5- PB-0-3- PC-0-2- P A - B - 0-15-P A - C - 0-1 and P B - C - 0-06 The smallest possible value of P AC-BC-CC is

The correct option is A 0.31 Total probability P(A∪ B∪ C)=1 P(A∪ B∪ C)=P(A)+P(B)+P(C) P(A∩ B) P(B∩ C) P(C∩ A) P(A∩ B∩ C), P(A∪ B∪ C)=0.5+0.3+ ...

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Standard XII

Mathematics

Properties of Set Operation

Let A, B, C b...

Question

Let A, B, C be three events in a probability space. Suppose that $P (A) = 0.5, P (B) = 0.3, P (C) = 0.2$ , $P (A \cap B) = 0.15, P (A \cap C) = 0.1$ and $P (B \cap C) = 0.06$ The smallest possible value of $P (A^{C} \cap B^{C} \cap C^{C})$ is

0.31

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Solution

The correct option is A $0.31$
Total probability $P (A \cup B \cup C) = 1$
$P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) - P (A \cap B \cap C), P (A \cup B \cup C) = 0.5 + 0.3 + 0.2 - 0.15 - 0.1 - 0.06 - 1, 1 = 1 - 0.31 - 1 (A \cap B \cap C), P (A \cap B \cap C) = 0.31$

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Similar questions

Let A, B, C be three events in a probability space. Suppose that P(A)=0.5, P(B)=0.3, P(C)=0.2,
$P (A \cap B) = 0.15, P (A \cap C) = 0.1$ and $P (B \cap C) = 0.06$
The maximum possible value of $P (A^{'} \cap B^{'} \cap C^{'})$ is,

Q. Let A, B, C be three events such that :

P (A) = 0.3, P (B) = 0.4, P (C) = 0.5, P (A \cap B^{^{'}}) = 0.2, P (B \cap C) = 0.3, P (A^{^{'}} \cap B^{^{'}} \cap C^{^{'}}) = 0.3

and

P (A \cap B | C^{^{'}}) = 0.1

. Find the value of

P (B^{^{'}} | C^{^{'}})

Q. The probabilities of three events

A, B

and

C

are

P (A) = 0.6, P (B) = 0.4, P (C) = 0.5

,also

P (A \cup B) = 0.8, P (A \cap C) = 0.3, P (A \cup B \cup C) \geq 0.85, P (A \cap B \cap C) = 0.2

and

P (B \cap C) = p_{1}

. Then

Q. The probabilities of three events

A, B

and

C

are

P (A) = 0.6, P (B) = 0.4

and

P (C) = 0.5

. If

P (A \cup B) = 0.8, P (A \cap C) = 0.3, P (A \cap B \cap C) = 0.2

and

P (A \cup B \cup C) \geq 0.85

0.04 k \leq P (B \cap C) \leq 0.07 k

. Find the value of

k

Q. Let

A, B

C

3

arbitrary events defined on a sample space

S

and if,

P (A) + P (B) + P (C) = \frac{1}{2}, P (A \cap B) + P (B \cap C) + P (C \cap A) = \frac{1}{4}

P (A \cap B \cap C) = \frac{1}{6}

, then the probability that exactly one of the three events occurs is

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Let A, B, C be three events in a probability space. Suppose that P(A)=0.5,P(B)=0.3,P(C)=0.2, P(A∩B)=0.15,P(A∩C)=0.1 and P(B∩C)=0.06 The smallest possible value of P(AC∩BC∩CC) is

The correct option is A 0.31Total probability P(A∪B∪C)=1P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)−P(A∩B∩C),P(A∪B∪C)=0.5+0.3+0.2−0.15−0.1−0.06−1,1=1−0.31−1(A∩B∩C),P(A∩B∩C)=0.31

Let A, B, C be three events in a probability space. Suppose that $P (A) = 0.5, P (B) = 0.3, P (C) = 0.2$ , $P (A \cap B) = 0.15, P (A \cap C) = 0.1$ and $P (B \cap C) = 0.06$ The smallest possible value of $P (A^{C} \cap B^{C} \cap C^{C})$ is