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 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,
$\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right)+\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{A})$
$=\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3}$
$=\frac{1}{3}+\frac{1}{6}$
$=\frac{1}{2}$
Similar Questions
If $P(A) = P(B) = x$ and $P(A \cap B) = P(A' \cap B') = \frac{1}{3}$, then $x = $
If $P(A) = P(B) = x$ and $P(A \cap B) = P(A' \cap B') = \frac{1}{3}$, then $x = $
If $A$ and $B$ are two events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{2}{3},$ then
If $A$ and $B$ are two events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{2}{3},$ then
$\mathrm{A}$ die is thrown. If $\mathrm{E}$ is the event $'$ the number appearing is a multiple of $3'$ and $F$ be the event $'$ the number appearing is even $^{\prime}$ then find whether $E$ and $F$ are independent ?
$\mathrm{A}$ die is thrown. If $\mathrm{E}$ is the event $'$ the number appearing is a multiple of $3'$ and $F$ be the event $'$ the number appearing is even $^{\prime}$ then find whether $E$ and $F$ are independent ?
Two persons $A$ and $B$ throw a (fair)die (six-faced cube with faces numbered from $1$ to $6$ ) alternately, starting with $A$. The first person to get an outcome different from the previous one by the opponent wins. The probability that $B$ wins is
Two persons $A$ and $B$ throw a (fair)die (six-faced cube with faces numbered from $1$ to $6$ ) alternately, starting with $A$. The first person to get an outcome different from the previous one by the opponent wins. The probability that $B$ wins is
- [KVPY 2014]
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is $0.05$ and that Ashima will qualify the examination is $0.10 .$ The probability that both will qualify the examination is $0.02 .$ Find the probability that Both Anil and Ashima will not qualify the examination.
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is $0.05$ and that Ashima will qualify the examination is $0.10 .$ The probability that both will qualify the examination is $0.02 .$ Find the probability that Both Anil and Ashima will not qualify the examination.