In a game of chance, the player randomly draws two balls at the SAME TIME from an urn containing 2 red balls and 5 black balls. Find the probability that the player draws exactly one red ball.
Can anyone pls help? Thanks!
2006-11-07 18:54:55 · 6 answers · asked by sky_blue 1 in Science & Mathematics ➔ Mathematics
Total number of balls: 7
Let's say the player draws each ball with each hand.
There can be 2 situations:
1)Player draws 1 red ball with left hand and 1 black ball with right hand: (2/7 * 5/6)
2)Player draws 1 red ball with right hand and 1 black ball with left hand: (5/7 * 2/6)
Therefore, probability of drawing exactly 1 red ball is (2/7 * 5/6) + (5/7 * 2/6)= 10/21
2006-11-07 19:05:28 · answer #1 · answered by ispakles 3 · 0⤊ 0⤋
Drawing two balls at the same time is the same as taking one ball at a time without replacing.
P(1 red ball,1 black ball)=2(2/7*5/6)
=2 X 10/42
=10/21
2006-11-08 03:00:34 · answer #2 · answered by A 150 Days Of Flood 4 · 1⤊ 0⤋
OK. This is a imple counting problem. How many ways are the to select 2 things from 7? How many of those have exaclty 1 red ball?
2006-11-08 03:06:05 · answer #3 · answered by Mich Ravera 3 · 0⤊ 0⤋
total no. of balls = 2+5 =7
probability of getting exactly 1 red ball(by applying combination)=2C1
now remaining ball= 1
so its probability=5C1
now probability of getting two balls= (2C1*5C1)/7C2
by solving this it will be
(2*5*2*1)/(7*6) = 10/21
2006-11-08 03:24:43 · answer #4 · answered by neeti 2 · 0⤊ 0⤋
As he draws exactly one red ball, it means he draws 1red and 1 black ball.
P = 2C1 * 5C1 / 7C2
= 2.5.2/ 7.6 = 10/21
2006-11-08 03:06:20 · answer #5 · answered by Amit K 2 · 1⤊ 0⤋
Total Combinations are:
(1R, 2R) (1R, 1B) (1R, 2B)...(1R, 5B) 6
(2R, 1B)..(2R, 5B) 5
(1B, 2B)..(1B,5B) 4
(2B, 3B)...(2B, 5B) 3
(3B,4B)(3B,5B) 2
(4B, 5B) 1
21 total combinations
10 one red ball combination.
The probability is then 10/21
2006-11-08 03:11:48 · answer #6 · answered by mo2lose 2 · 0⤊ 0⤋