Suppose that P(A)=0.3 and P(B)=0.25 andP(A∩B)=0.1. What is P(B∣A(complement))?

Answer:P(B∣A′)=0.5Explanation:We use the definition of conditional probability:P(X∣Y)=P(X∩Y)P(Y)along withP(X∩Y′)=P(X)−p(X∩Y)From this we get;P(B∣A′)=P(B∩A′)P(A′)And P(B∩A′)=P(B)−P(B∩A)We are given that P(A)=0.3,P(B)=0.25 and P(A∩B)=0.1, so:P(A)=0.3⇒P(A′)=0.7P(B∩A′)=0.25−0.1=0.15P(B∣A′)=0.150.3=0.5

"Suppose that P(A)=0.3 and P(B)=0.25 andP(A∩B)=0.1. What is" was answered by our community

jkminzeszjt

Answered question

2022-02-15

Suppose that

P (A) = 0.3 and P (B) = 0.25 and P (A \cap B) = 0.1

. What is

P (B ∣ A (complement))

Answer & Explanation

Utskoldendv3

Beginner2022-02-16Added 9 answers

Answer:

P (B ∣ A^{'}) = 0.5

Explanation:
We use the definition of conditional probability:

P (X ∣ Y) = \frac{P (X \cap Y)}{P (Y)}

along with

P (X \cap Y^{'}) = P (X) - p (X \cap Y)

From this we get;

P (B ∣ A^{'}) = \frac{P (B \cap A^{'})}{P (A^{'})}

And

P (B \cap A^{'}) = P (B) - P (B \cap A)

We are given that

P (A) = 0.3, P (B) = 0.25 and P (A \cap B) = 0.1

, so:

P (A) = 0.3 \Rightarrow P (A^{'}) = 0.7

P (B \cap A^{'}) = 0.25 - 0.1 = 0.15

P (B ∣ A^{'}) = \frac{0.15}{0.3} = 0.5

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school probability

Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an LG Smartphone survey). If 8 adult smartphone users are randomly selected, find the probability that exactly 6 of them use their smartphones in meetings or classes?
Write formula for the sequence of -4, 0, 8, 20, 36, 56, 80, where the order of f(x) is 0, 1, 2, 3, 4, 5, 6 respectively
A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment.n=20,
p=0.7,
x=19
P(19)=
In binomial probability distribution, the dependents of standard deviations must includes.
a) all of above.
b) probability of q.
c) probability of p.
d) trials.
The probability that a man will be alive in 25 years is 3/5, and the probability that his wifewill be alive in 25 years is 2/3
Determine the probability that both will be alive
How many different ways can you make change for a quarter??
(Different arrangements of the same coins are not counted separately.)
One hundred people line up to board an airplane that can accommodate 100 passengers. Each has a boarding pass with assigned seat. However, the first passenger to board has misplaced his boarding pass and is assigned a seat at random. After that, each person takes the assigned seat. What is the probability that the last person to board gets his assigned seat unoccupied?
A) 1
B) 0.33
C) 0.6
D) 0.5
The value of $(243)^{- \frac{2}{5}}$ is _______.
A)9
B) $\frac{1}{9}$
C) $\frac{1}{3}$
D)0
1 octopus has 8 legs. How many legs does 3 octopuses have?
A) 16
B 24
C) 32
D) 14
From a pack of 52 cards, two cards are drawn in succession one by one without replacement. The probability that both are aces is...
A pack of cards contains $4$ aces, $4$ kings, $4$ queens and $4$ jacks. Two cards are drawn at random. The probability that at least one of these is an ace is A $\frac{9}{20}$ B $\frac{3}{16}$ C $\frac{1}{6}$ D $\frac{1}{9}$
You spin a spinner that has 8 equal-sized sections numbered 1 to 8. Find the theoretical probability of landing on the given section(s) of the spinner. (i) section 1 (ii) odd-numbered section (iii) a section whose number is a power of 2. [4 MARKS]
If A and B are two independent events such that $P (A) > 0.5, P (B) > 0.5, P (A \cap \bar{B}) = \frac{3}{25} P (\bar{A} \cap B) = \frac{8}{25}$ , then the value of $P (A \cap B)$ is
A) $\frac{12}{25}$
B) $\frac{14}{25}$
C) $\frac{18}{25}$
D) $\frac{24}{25}$
The unit of plane angle is radian, hence its dimensions are
A) $[M^{0} L^{0} T^{0}]$
B) $[M^{1} L^{- 1} T^{0}]$
C) $[M^{0} L^{1} T^{- 1}]$
D) $[M^{1} L^{0} T^{- 1}]$
Clinical trial tests a method designed to increase the probability of conceiving a girl. In the study, 340 babies were born, and 289 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born?

Didn't find what you were looking for?