(D) We know that $P(\bar{A}) = 1 - P(A)$ and $P(\bar{B}) = 1 - P(B).$Therefore,$P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B)).

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(D) We know that $P(\bar{A}) = 1 - P(A)$ and $P(\bar{B}) = 1 - P(B).$Therefore,$P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B)).$Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B).$Substituting the given values: $0.8 = P(A) + P(B) - 0.3.$Thus,$P(A) + P(B) = 0.8 + 0.3 = 1.1.$Finally,$P(\bar{A}) + P(\bar{B}) = 2 - 1.1 = 0.9.$

Class 11 Mathematics Probability Odds in favor and odds against, Addition theorem on probability

Medium

If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $

A
$0.3$
B
$0.5$
C
$0.7$
D
$0.9$

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Class 11 Questions

Class 11

Mathematics Questions

Chapter

Probability

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Odds in favor and odds against, Addition theorem on probability

Two dice are thrown simultaneously. The probability that the sum is odd or less than $7$ or both is:

Medium

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If $A$ and $B$ are two independent events,then $P(A \cup B) = $

Medium

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$A$ card is drawn at random from a pack of $52$ cards. The probability of this card being a red card or a queen is:

Easy

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If $A$ and $B$ are two events such that $P(A \cup B) = \frac{5}{6}$,$P(A \cap B) = \frac{1}{3}$ and $P(\bar{B}) = \frac{1}{3}$,then $P(A) = $

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$A$ card is drawn from a pack of $52$ cards. $A$ gambler bets that it is a spade or an ace. What are the odds against his winning this bet?

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If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $

Similar Questions

Two dice are thrown simultaneously. The probability that the sum is odd or less than $7$ or both is:

If $A$ and $B$ are two independent events,then $P(A \cup B) = $

$A$ card is drawn at random from a pack of $52$ cards. The probability of this card being a red card or a queen is:

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