# Impossible... I cant get this.. If any one else can There a genius..

• March 5th 2008, 10:46 PM
wikji
Impossible... I cant get this.. If any one else can There a genius..
7. A drawer contains four red socks and five blue socks.
(a) Three socks are drawn one at a time and then put back before the next
selection. Determine the probability that
(i) exactly two red socks are selected
(ii) at least two red socks are selected
(b) Repeat part (a) without replacement

iia) 5over14 iib) 17over42

(over is a fraction)
• March 5th 2008, 11:12 PM
TheEmptySet
Quote:

Originally Posted by wikji
7. A drawer contains four red socks and five blue socks.
(a) Three socks are drawn one at a time and then put back before the next
selection. Determine the probability that
(i) exactly two red socks are selected
(ii) at least two red socks are selected
(b) Repeat part (a) without replacement

iia) 5over14 iib) 17over42

(over is a fraction)

to pick one red sock

$\frac{\binom{4}{1}}{\binom{9}{1}}$

to pick one blue sock

$\frac{\binom{5}{1}}{\binom{9}{1}}$

Since we are replacing the socks the events are independant so we multiply the probibilites together

$\frac{\binom{4}{1}}{\binom{9}{1}}\cdot \frac{\binom{4}{1}}{\binom{9}{1}} \cdot \frac{\binom{5}{1}}{\binom{9}{1}}=\frac{80}{729}$

since there are three different ways we can draw 2 red and 1 blue {r,r,b} or {r,b,r} or {b,r,r} we multiply our prob by 3

$\frac{80}{729} \cdot 3 =\frac{80}{243}$

for two use the same process to find the prob of 3 red socks. then add the answer from part 1 to this answer.

Without replacement

$\frac{\binom{4}{2} \binom{5}{1}}{\binom{9}{3}}=\frac{5}{14}$

use the same as above for the 2nd part