A person writes 4 letters and addresses 4 envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelopes, is
A$\frac{1}{4}$
B$\frac{11}{14}$
C$\frac{15}{24}$
D$\frac{23}{24}$
Download our app for free and get started
D$\frac{23}{24}$
4 letter can be placed in 4 envelopes in 4! ways = 24ways
Now, there is only one method, by which all the letters are placed in the right envelope.
P(all letters are placed in the envelopes) $=\frac{1}{24}$
P(all letters are not placed in the right envelopes) = 1 - P(all letters are placed in the right envelopes)
$=1-\frac{1}{24}=\frac{23}{24}$
Download our app
and get started for free
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.*
Choose the correct answer from the given four options.
A box has 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective?
Choose the correct answer from the given four options.
If $\text{P}(\text{A})=\frac{3}{10},\text{P}(\text{B})=\frac{2}{5}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{5},$ then $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)+\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ equas:
If A and B are two events such that $\text{P}(\text{A}|\text{B})=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9},$ then p =