(A) Since $A$ and $B$ are independent events,$P(A \cap B) = P(A) \times P(B) = 0.3 \times 0.6 = 0.18$. We need to find $P(\text{neither } A \text{ nor

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(A) Since $A$ and $B$ are independent events,$P(A \cap B) = P(A) \times P(B) = 0.3 \times 0.6 = 0.18$. We need to find $P(\text{neither } A \text{ nor } B)$,which is $P(A' \cap B')$. By De Morgan's Law,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$. Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.6 - 0.18 = 0.72$. Therefore,$P(A' \cap B') = 1 - 0.72 = 0.28$.

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

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Given two independent events $A$ and $B$ such that $P(A) = 0.3$ and $P(B) = 0.6$. Find $P(\text{neither } A \text{ nor } B)$.

A
$0.28$
B
$0.32$
C
$0.42$
D
$0.72$

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Probability

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Odds in favor and odds against, Addition theorem on probability

$A$ random variable $X$ has the probability distribution:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,then $P(E \cup F)$ is

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One of the two events must occur. If the chance of one is $\frac{2}{3}$ of the other,then the odds in favour of the other are

Medium

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The odds against an event are $5 : 2$ and the odds in favor of another independent event are $6 : 5$. What is the probability that at least one of the events occurs?

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Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find $P(A \cup B)$.

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One card is drawn from a pack of $52$ cards. The probability that it is a king or a diamond is

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$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

Given two independent events $A$ and $B$ such that $P(A) = 0.3$ and $P(B) = 0.6$. Find $P(\text{neither } A \text{ nor } B)$.

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$A$ random variable $X$ has the probability distribution:$X$$1$$2$$3$$4$$5$$6$$7$$8$$P(X)$$0.15$$0.23$$0.12$$0.10$$0.20$$0.08$$0.07$$0.05$For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,then $P(E \cup F)$ is

One of the two events must occur. If the chance of one is $\frac{2}{3}$ of the other,then the odds in favour of the other are

The odds against an event are $5 : 2$ and the odds in favor of another independent event are $6 : 5$. What is the probability that at least one of the events occurs?

Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find $P(A \cup B)$.

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$A$ random variable $X$ has the probability distribution:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,then $P(E \cup F)$ is