Question 480257: A bag contains 6 red, 4 white and 8 blue balls. If 3 balls are drawn at random, determine the probability p that at least 1 is red.
Here is how I did it:
P(at least 1 Red) = ?
3 drawings
Probability that 1 is Red in any trial = (1/18)X6 = 1/3
Probability that 1 is NOT Red in any trial = q = 1 - p = 1 - 1/3 = 2/3
Probability that 1 is NOT Red in 3 drawings = (2/3)X(2/3)X(2/3) = 8/27.
Hence the probability of getting at least 1 Red in 3 drawings = 1-(8/27)= 19/27 = 0.70
However, the textbook's answer and the solving method are different:
Probability that none is Red = [(4+8)C3]/18C3 = 55/204
Hence the probability that at least 1 is Red = 1-(55/204) = 149/204 = 0.73
Could you tell me why is my method wrong? I used the method because a similar problem in my textbook was solved using it. If you like here is the problem I am referring to:
What is the probability of getting at least 1 ace in 2 throws of a die?
The probability of getting an ance in any single throw = 1/6.
The probability of NOT getting an ace in any single throw = 1-(1/6)=5/6.
The probability of not getting an ace in 2 throws = (5/6)X(5/6)=25/36.
Hence the probability of getting at least 1 ace in 2 throws = 1-(25/36)=11/36.
What is the difference between the two types of problems and why my method did not work?
Thank you so much.
Have a nice evening.
Respectfully,
Ivanka
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
If you were to reach in the bag, pull out 1 ball, check to see if it is red or not, THEN REPLACE IT and repeat the experiment twice more, then your method would produce the correct probability, namely 1 minus the binomial distribution of zero successes in 3 trials where the success of one trial is 1/3. Notice that you have to do it with replacement because if you don't, the probability of a red changes on each trial (either by reducing the denominator of the single trial probability fraction by one if you DON'T get a red, or reducing both the numerator AND denominator by one if you DO get a red). Note that the binomial distribution works for the dice game because you are always replacing -- that is you never remove any spots from your dice before rolling them another time.
However, the experiment described is performed by reaching in the bag, grabbing 3 marbles and checking to see if any of them are red. Different experiment -- different results. Here you need the number of ways the single outcome could be zero reds, namely 12 choose 3, divided by the number of ways to draw 3 balls at once, namely 18 choose 3 and then subtract that result from 1.
John
My calculator said it, I believe it, that settles it
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