The probability distribution of a random variable X is given below :
| X | 4k | $\frac{30}{7}k$ | $\frac{32}{7}k$ | $\frac{34}{7}k$ | $\frac{36}{7}k$ | $\frac{38}{7}k$ | $\frac{40}{7}k$ | 6k |
|---|---|---|---|---|---|---|---|---|
| P(X) | $\frac{2}{15}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{1}{5}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{1}{5}$ | $\frac{1}{15}$ |
If E(X) = $\frac{263}{15}$, then P(X < 20) is equal to :
A bag contains 10 balls out of which $k$ are red and $(10-k)$ are black, where $0 \leq k \leq 10$. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:
From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is
Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A . Then a ball is randomly drawn from the bag A . If the probability, that the ball drawn is white, is $\frac{\mathrm{p}}{\mathrm{q}}, \operatorname{gcd}(\mathrm{p}, \mathrm{q})=1$, then $\mathrm{p}+\mathrm{q}$ is equal to
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