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If 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 $\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=$
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:
Choose the correct answer from the given four options.
Let A and B be two events such that $\text{P}(\text{A})=\frac{3}{8},\text{P}({\text{B}})=\frac{5}{8}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{4}.$Then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}\Big(\frac{\text{A'}}{\text{B}}\Big)$ is equal to:
If X is a binomial variate with parameters n and p, where 0 < p < 1 such that $\frac{\text{P(X = r)}}{\text{P(X = n - r})}$ is independent of n and r, then p equals:
A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is: