$A$ bag contains $4$ white and $3$ red balls. Two draws of one ball each are made without replacement. Then the probability that both the balls are red is
- A$\frac{1}{7}$
- B$\frac{2}{7}$
- C$\frac{3}{7}$
- D$\frac{4}{7}$
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View Solution$A$ bag contains $6$ red, $2$ white and $8$ blue balls. Three balls are drawn at random from the bag. Match the items of List-$I$ with those of List-$II$.
$A$. Probability that none of the balls is white $I$. $\frac{1}{70}$ $B$. Probability of getting $2$ white and $1$ blue ball $II$. $\frac{6}{35}$ $C$. Probability of getting $2$ blue and $1$ white ball $III$. $\frac{13}{20}$ $D$. Probability of getting $1$ red, $1$ white and $1$ blue ball $IV$. $\frac{1}{10}$
| $A$. Probability that none of the balls is white | $I$. $\frac{1}{70}$ |
| $B$. Probability of getting $2$ white and $1$ blue ball | $II$. $\frac{6}{35}$ |
| $C$. Probability of getting $2$ blue and $1$ white ball | $III$. $\frac{13}{20}$ |
| $D$. Probability of getting $1$ red, $1$ white and $1$ blue ball | $IV$. $\frac{1}{10}$ |
Out of the first $5$ consecutive natural numbers,if two different numbers $x$ and $y$ are chosen at random,then the probability that $x^4 - y^4$ is divisible by $5$ is
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