Advertisements
Advertisements
Question
Events A and Bare such that P(A) = `1/2`, P(B) = `7/12` and `P(barA ∪ barB) = 1/4`. Find whether the events A and B are independent or not.
Solution
Given, P(A) = `1/2`, P(B) = `7/12`
And `P(barA ∪ barB) = 1/4`
For A and B are independent
P(A ∩ B) = P(A).P(B) ...(i)
Now, `P(barA ∪ barB)` = P(A ∩ B)
⇒ `P(barA ∪ barB)` = 1 – P(A ∩ B)
⇒ P(A ∩ B) = `1 - P(barA ∪ barB)`
⇒ P(A ∩ B) = `1 - 1/4 = 3/4` ...(ii)
Now, P(A).P(B) = `1/2 xx 7/12 = 7/24` ...(iii)
Since from equations (ii) and (ill)
P(A ∩ B) ≠ P(A).P(B)
Therefore, events A and B are not independent.
APPEARS IN
RELATED QUESTIONS
A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.
If A and B are two events such that `P(A) = 1/4, P(B) = 1/2 and P(A ∩ B) = 1/8`, find P (not A and not B).
Two events, A and B, will be independent if ______.
If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability `1/2`).
A speaks the truth in 60% of the cases, while B is 40% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact?
The odds against a husband who is 55 years old living till he is 75 is 8: 5 and it is 4: 3 against his wife who is now 48, living till she is 68. Find the probability that the couple will be alive 20 years hence.
The odds against a husband who is 55 years old living till he is 75 is 8: 5 and it is 4: 3 against his wife who is now 48, living till she is 68. Find the probability that at least one of them will be alive 20 years hence.
The probability that a student X solves a problem in dynamics is `2/5` and the probability that student Y solves the same problem is `1/4`. What is the probability that
- the problem is not solved
- the problem is solved
- the problem is solved exactly by one of them
Two hundred patients who had either Eye surgery or Throat surgery were asked whether they were satisfied or unsatisfied regarding the result of their surgery
The follwoing table summarizes their response:
| Surgery | Satisfied | Unsatisfied | Total |
| Throat | 70 | 25 | 95 |
| Eye | 90 | 15 | 105 |
| Total | 160 | 40 | 200 |
If one person from the 200 patients is selected at random, determine the probability that the person was satisfied given that the person had Throat surgery.
A bag contains 3 yellow and 5 brown balls. Another bag contains 4 yellow and 6 brown balls. If one ball is drawn from each bag, what is the probability that, the balls are of different color?
Bag A contains 3 red and 2 white balls and bag B contains 2 red and 5 white balls. A bag is selected at random, a ball is drawn and put into the other bag, and then a ball is drawn from that bag. Find the probability that both the balls drawn are of same color
A bag contains 3 red and 5 white balls. Two balls are drawn at random one after the other without replacement. Find the probability that both the balls are white.
Solution: Let,
A : First ball drawn is white
B : second ball drawn in white.
P(A) = `square/square`
After drawing the first ball, without replacing it into the bag a second ball is drawn from the remaining `square` balls.
∴ P(B/A) = `square/square`
∴ P(Both balls are white) = P(A ∩ B)
`= "P"(square) * "P"(square)`
`= square * square`
= `square`
A family has two children. Find the probability that both the children are girls, given that atleast one of them is a girl.
Solve the following:
Let A and B be independent events with P(A) = `1/4`, and P(A ∪ B) = 2P(B) – P(A). Find P(B)
Solve the following:
Find the probability that a year selected will have 53 Wednesdays
Solve the following:
For three events A, B and C, we know that A and C are independent, B and C are independent, A and B are disjoint, P(A ∪ C) = `2/3`, P(B ∪ C) = `3/4`, P(A ∪ B ∪ C) = `11/12`. Find P(A), P(B) and P(C)
Solve the following:
A machine produces parts that are either good (90%), slightly defective (2%), or obviously defective (8%). Produced parts get passed through an automatic inspection machine, which is able to detect any part that is obviously defective and discard it. What is the quality of the parts that make it throught the inspection machine and get shipped?
If A and B′ are independent events then P(A′ ∪ B) = 1 – ______.
Let A and B be two independent events. Then P(A ∩ B) = P(A) + P(B)
Three events A, B and C are said to be independent if P(A ∩ B ∩ C) = P(A) P(B) P(C).
Refer to Question 1 above. If the die were fair, determine whether or not the events A and B are independent.
If A and B are two events such that P(A) = `1/2`, P(B) = `1/3` and P(A/B) = `1/4`, P(A' ∩ B') equals ______.
If A and B are two independent events with P(A) = `3/5` and P(B) = `4/9`, then P(A′ ∩ B′) equals ______.
If A, B and C are three independent events such that P(A) = P(B) = P(C) = p, then P(At least two of A, B, C occur) = 3p2 – 2p3
If P(A) = `3/5` and P(B) = `1/5`, find P(A ∩ B), If A and B are independent events.
Let A and B be independent events P(A) = 0.3 and P(B) = 0.4. Find P(A ∩ B)
Let EC denote the complement of an event E. Let E1, E2 and E3 be any pairwise independent events with P(E1) > 0 and P(E1 ∩ E2 ∩ E3) = 0. Then `"P"(("E"_2^"C" ∩ "E"_3^"C")/"E"_1)` is equal to ______.
The probability of the event A occurring is `1/3` and of the event B occurring is `1/2`. If A and B are independent events, then find the probability of neither A nor B occurring.
