Choose the correct answer from the given four options. A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is:
A$\frac{45}{196}$
B$\frac{135}{392}$
C$\frac{15}{56}$
D$\frac{15}{29}$
Download our app for free and get started
C$\frac{15}{56}$
Probability of getting exactly one red (R) ball
$=\text{P}_{\text{R}}\cdot\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\bar{\text{R}}}+\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\text{R}}\cdot\text{P}_{\bar{\text{R}}}+\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\text{R}}$
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Assume that in a family, each chold is equally likely to be a boy or a girl. A family with tree cgildren is chosen at random. Tere probability that the eldest child is a girl given that the family has at least oe girl.
A pack of playing cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, then the probability thatthe missing card is black, is:
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?
Choose the correct answer from the given four options$.A$ and $B$ are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4},$ respectively. If the probability of their making a common error is$, \frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is:
In a college 30% students fail in Physics, 25% fail in Mathenatics and 10% fail in both. One student is chosen at random. The probability that she fails in Physics if she failed in Mathematics is.
Choose the correct answer from the given four options.
If A and B are two events such that $\text{P}(\text{A})=\frac{1}{2},\text{P}(\text{B})=\frac{1}{3},$ $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4},$ then $\text{P}(\text{A}'\cap\text{B}')$ equals: