Question

A box contains 10 red marbles,20 blue marbles and 30 green marbles.5 marbles are drawn from the box,what is the probability that
1))all are blue
2))atleast one is green

Answer 1

@hartnn

Answer 2

You first have to know is probability percents or fractions?

Answer 3

^wrong

Answer 4

The question implies that more than one marble is drawn. How many marbles are drawn? Are the marbles drawn without replacement?

Answer 5

some typo here, probably a cut and past problem

Answer 6

5,fixed

Answer 7

Can we assume that 5 marbles are drawn without replacement?

Answer 8

yes

Answer 9

number of cases when all are blue 10c5
number of cases=50c5 .. divide.

Answer 10

20c5 *

Answer 11

\[P(5\ blue)=\frac{\left(\begin{matrix}20 \\ 5\end{matrix}\right)\left(\begin{matrix}40 \\ 0\end{matrix}\right)}{\left(\begin{matrix}60 \\ 5\end{matrix}\right)}\]

Answer 12

part 2-- 1- (prob of no green marbles)

Answer 13

got it! thanks!

Answer 14

all are blue
\[\frac{20}{60}\times \frac{19}{59}\times \frac{18}{58}\times \frac{17}{57}\times \frac{16}{56}\]

Answer 15

at least one is green
compute the probability that none are green, subtract from 1

Answer 16

\[P(zero\ green)=\frac{5!55!}{60!}\]
\[P(at\ least\ one\ green)=1-\frac{5!55!}{60!}=?\]