Probability of solving specific problem independently by $A$ and $B$ are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that exactly one of them problem
Probability of solving the problem by $\mathrm{A},\, \mathrm{P}(\mathrm{A})=\frac{1}{2}$
Probability of solving the problem by $\mathrm{B}, \,\mathrm{P}(\mathrm{B})=\frac{1}{3}$
since the problem is solved independently by $A$ and $B$,
$\therefore $ $\mathrm{P}(\mathrm{AB})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$
$P(A^{\prime})=1-P(A)=1-\frac{1}{2}=\frac{1}{2}$
$P(B^{\prime})=1-P(B)=1-\frac{1}{3}=\frac{2}{3}$
Probability that exactly one of them solves the problem is given by,
$\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right)+\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{A})$
$=\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3}$
$=\frac{1}{3}+\frac{1}{6}$
$=\frac{1}{2}$
For two given events $A$ and $B$, $P\,(A \cap B) = $
If $A$ and $B$ are two events such that $P\left( {A \cup B} \right) = P\left( {A \cap B} \right)$, then the incorrect statement amongst the following statements is
Fill in the blanks in following table :
$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
$0.5$ | $0.35$ | ......... | $0.7$ |
Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined $P ( A )=0.5$, $ P ( B )=0.4$, $P ( A \cap B )=0.8$
A card is drawn at random from a pack of cards. The probability of this card being a red or a queen is