Answer to Question #270438 in Statistics and Probability for Light

Question

On a certain day, the district jail had 190 prisoners. Of these, 130 were accused of felonies and 121 were accused of misdemeanors. If a prisoner was selected at random from the jail what is the probability that he was accused of both crimes? (4 marks) A box contains 6 red, 5 black and 4 green beads. If two beads are randomly selected, one at a time with replacement, what is the probability that a red and a black bead will be selected?

Answer 1

(a) Let A - set of prisoners accused of felonies, B - set of prisoners accused of misdemeanors, then $|A∪B|=|C| =190$

$|A∪B|=|A|+|B|-|A\cap B|\implies |A\cap B|=|A|+|B|-|A∪B|=130+121-190=61$

Then for randomly selected prisoner probability of being accused for both crimes is ${\frac {61} {190}}\approx0.321$

(b) The probability of getting one red and one black bead is equal to probability of getting red than black plus getting black then red. Let A be "one red and one black bead", then

$P(A)=P(rb)+P(br)={\frac 6 {15}}*{\frac 5 {15}}+{\frac 5 {15}}*{\frac 6 {15}}={\frac {60} {225}}={\frac 4 {15}}$