Choose the correct answer from the given four options.
You are given that A and B are two events such that $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then P(A) equals:
You are given that A and B are two events such that $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then P(A) equals:
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1View SolutionA and B are two events such that P(A) = 0.25 and P(B) = 0.50. The probability pf both happening together is 0.14. The probability of both A and B hot happening is.
- 2View SolutionChoose the correct answer from the given four options.Which one is not a requirement of a binomial distribution?
- 3View SolutionA person writes 4 letters and addresses 4 envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelopes, is
- 4Choose the correct answer from the given four options.If $\text{P}(\text{A})=0.4,\text{P}(\text{B})=0.8$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6,$ then $\text{P}(\text{A}\cup\text{B})$ is equal to:View Solution
- 5View SolutionA fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is:
- 6View SolutionA coin is tossed 4 times. The probability that at least one head turns up is:
- 7View SolutionA rifleman is firing at a distant target and has only 10% chance of hiting it. the least number of round he must fire in order to have more than 50% chance of hitting it at least once is:
- 8View SolutionOut of 30 consecutive integers, 2 are chosen at random. The probability that their sum is odd, is
- 9If A and B are such that $\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$ and $\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3},$ then $\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=$View Solution
- 10View SolutionA fair coin is tossed 100 times. The probability of getting tails an odd nimber of times is: