Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
BITSAT2023MathematicsProbabilityActual

If E and F are events such that 0 < P ( F ) < 1 , then

Options

  1. AP(E F)+P( E F)=1
  2. BP(E F)+P(E F )=1
  3. CP( E F)+P(E F )=1
  4. DP(E F )+P( E F )=0

Correct answer

A. P(E F)+P( E F)=1

Step-by-step solution

P(E F)+P( E F) = P(E F)+P( E F) P(F) = P((E E ) F) P(F) = P(F) P(F) =1

Practice Probability on Quantrex Academy →

More from Probability

The probability of getting 10 in a single throw of three fair dice is 2024In a binomial distribution, the mean is 4 and variance is 3 . Then, its mode is 2024The probability that certain electronic component fails when first used is 0.10 . If it does not fail immediately, the probability that it lasts for one year is 0.99 . The probabil 2024Given below is the distribution of a random variable X If = P (X 3) and = P (X 2) , then : = 2024A book contains 1000 pages. A page is chosen at random. The probability that the sum of the digits of the marked number on the page is equal to 9 , is 2024For two events A and B , if P ( A )= P ( A B )= 1 4 and P ( B A )= 1 2 , then which of the following is not true? 2024If A and B are mutually exclusive events and if P ( B )= 1 3 , P ( A B )= 13 21 , then P ( A ) is equal to 2023A coin is tossed twice. Then, the probability that atleast one tail occurs is 2023 Full Probability list All BITSAT PYQs