Another question which seems confusing,
There are 4 red socks and 2 blue socks in a drawer. 2 socks are removed at random. What is the probability of obtaining?
a. two red socks b. one of each color?
Cheers.
Another question which seems confusing,
There are 4 red socks and 2 blue socks in a drawer. 2 socks are removed at random. What is the probability of obtaining?
a. two red socks b. one of each color?
Cheers.
Another way is this: Originally there are 6 socks in the drawer and 4 of them are red: the probability of getting a red sock the first time is 4/6= 2/3.
Now, there are 5 socks left in the drawer and 3 of them are red. The probability of getting a red sock now is 3/5. The probability of both of those things happening is the product: (2/3)(3/5).
You can do the same thing with "one of each color":
Originally there are 6 socks in the drawer and 2 of them are blue. What is the probability of getting a blue sock? If that happens there will be 5 socks left and 4 of them are red. What is the probability of getting a red sock? Multiplying those together gives the probability of "blue sock, red sock" in that order. In order to find the probability of "one of each color" you must also find the probability of "red sock, blue sock" in that order- but that should be easy now.
Hello, ilovemaths!
There are: .$\displaystyle {6\choose2} \:=\:15$ possible outcomes.There are 4 red socks and 2 blue socks in a drawer.
Two socks are removed at random.
There are: .$\displaystyle {4\choose2} \:=\:6$ ways to draw two red socks.What is the probability of obtaining (a) two red socks?
Therefore: .$\displaystyle P(\text{2 red}) \:=\:\frac{6}{15} \:=\:\frac{2}{5}$
There are: .$\displaystyle {4\choose 1} \:=\:4$ ways to draw one red sock.b. one of each color?
There are: .$\displaystyle {2\choose1} \:=\:2$ ways to draw one blue sock.
. . Hence, there are:] .$\displaystyle 4\cdot2 \:=\:8$ ways to draw one of each color.
Therefore: .$\displaystyle P(\text{one of each color}) \:=\:\frac{8}{15}$