A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing $15$ oranges out of which $12$ are good and $3$ are bad ones will be approved for sale.

Let $A, B,$ and $C$ be the respective events that the first, second, and the third drawn orange is good.

Therefore, probability that first drawn orange is good, $\mathrm{P}(\mathrm{A})=\frac{12}{15}$

The oranges are not replaced.

Therefore, probability of getting second orange good, $\mathrm{P}(\mathrm{B})=\frac{11}{14}$

Similarly, probability of getting third orange good, $\mathrm{P}(\mathrm{C})=\frac{10}{13}$

The box is approved for sale, if all the three oranges are good.

Thus, probability of getting all the oranges good $=\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}=\frac{44}{91}$

Therefore, the probability that the box is approved for sale is $\frac{44}{91}$.

Let two fair six-faced dice $A$ and $B$ be thrown simultaneously. If $E_1$ is the event that die $A$ shows up four, $E_2 $ is the event that die $B$ shows up two and $E_3$ is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true $?$

- [JEE MAIN 2016]

$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P (A$ or $B).$

If $A$ and $B$ are two independent events such that $P(A) > 0.5,\,P(B) > 0.5,\,P(A \cap \bar B) = \frac{3}{{25}},\,P(\bar A \cap B) = \frac{8}{{25}}$ , then $P(A \cap B)$ is

India plays two matches each with West Indies and Australia. In any match the probabilities of India getting point $0, 1$ and $2$ are $0.45, 0.05$ and $0.50$ respectively. Assuming that the outcomes are independents, the probability of India getting at least $7$ points is

- [IIT 1992]

Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined $P ( A )=0.5$, $ P ( B )=0.7$, $P ( A \cap B )=0.6$