English

In an examination, 30% of students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subject I and subject II. A student is selected at random

Advertisements
Advertisements

Question

In an examination, 30% of students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subject I and subject II. A student is selected at random, what is the probability that the student has failed in exactly one subject?

Sum
Advertisements

Solution

Let A be the event that the student failed in Subject I
B be the event that the student failed in Subject II

Then P(A) = 30% = `30/100`

P(B) = 20% = `20/100`

And P(A ∩ B) = 10% = `10/100 `

P (student failed in exactly one subject)
= P(A) + P(B) – 2P(A ∩ B)

= `30/100+20/100-2(10/100)`

= `30/100`

= 0.30

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Probability - Exercise 7.4 [Page 108]

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

In a game, a man wins Rs 5 for getting a number greater than 4 and loses Rs 1 otherwise, when a fair die is thrown. The man decided to thrown a die thrice but to quit as and when he gets a number greater than 4. Find the expected value of the amount he wins/loses


Determine P(E|F).

Two coins are tossed once, where 

E: no tail appears, F: no head appears


Determine P(E|F).

A die is thrown three times,

E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses


A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5} Find P (E|G) and P (G|E)


Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.


An urn contains 4 black, 5 white, and 6 red balls. Two balls are drawn one after the other without replacement, What is the probability that at least one ball is black?


Two cards are drawn one after the other from a pack of 52 cards without replacement. What is the probability that both the cards drawn are face cards?


Can two events be mutually exclusive and independent simultaneously?


A problem in Mathematics is given to three students whose chances of solving it are `1/3, 1/4` and `1/5`. What is the probability that the problem is solved?


Choose the correct alternative:

Let A and B be two events such that `"P"(bar ("A" ∪ "B")) = 1/6, "P"("A" ∩ "B") = 1/4` and `"P"(bar"A") = 1/4`. Then the events A and B are


The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is ______ 


If two balls are drawn from a bag containing 3 white, 4 black and 5 red balls. Then, the probability that the drawn balls are of different colours is:


For a biased dice, the probability for the different faces to turn up are

Face 1 2 3 4 5 6
P 0.10 0.32 0.21 0.15 0.05 0.17

The dice is tossed and it is told that either the face 1 or face 2 has shown up, then the probability that it is face 1, is ______.


If A and B are two events such that `P(A/B) = 2 xx P(B/A)` and P(A) + P(B) = `2/3`, then P(B) is equal to ______.


Read the following passage:

Recent studies suggest the roughly 12% of the world population is left-handed.

Depending upon the parents, the chances of having a left-handed child are as follows:

A :  When both father and mother are left-handed:
Chances of left-handed child is 24%.
B :  When father is right-handed and mother is left-handed:
Chances of left-handed child is 22%.
C :  When father is left-handed and mother is right-handed:
Chances of left-handed child is 17%.
D :  When both father and mother are right-handed:
Chances of left-handed child is 9%.

Assuming that P(A) = P(B) = P(C) = P(D) = `1/4` and L denotes the event that child is left-handed.

Based on the above information, answer the following questions:

  1. Find `P(L/C)` (1)
  2. Find `P(overlineL/A)` (1)
  3. (a) Find `P(A/L)` (2)
    OR
    (b) Find the probability that a randomly selected child is left-handed given that exactly one of the parents is left-handed. (2)

Which complement form is correct when \[P(B)\neq 0\]?


Which statement correctly describes the numerator in \[P(A\mid B)=\frac{P(A\cap B)}{P(B)}\]?


Which sequence correctly applies a given condition when finding a conditional probability?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×