Question 983386: A bag has 7 balls: a green, black, blue, brown, pink, red, and orange one. If 6 balls are removed from the bag at random, what is the probability that a blue ball has been removed?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let's say we had 6 slots A through F
7 choices for slot A
6 choices for slot B
5 choices for slot C
4 choices for slot D
3 choices for slot E
2 choices for slot F
7*6*5*4*3*2 = 5040
If order mattered, then there would be 5040 ways to pick the six balls. However, order doesn't matter so we divide by 6! = 720 to get 5040/720 = 7
There are 7 ways to pick the six balls (out of a pool of 7), where again, order doesn't matter.
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If we reserve a slot, say slot A, for a blue ball, then there are 5 slots left and 6 items to fill those slots.
Using the same idea above, we can take a shortcut using the combination formula to say 6 C 5 = 6
There are 6 ways to pick a set of balls where one is definitely blue.
This is out of 7 ways total to pick the balls (we don't care if we get blue or not here)
That's why the answer is 6/7
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Alternative Method:
Think of it in a reverse way. Think of it as "what is the probability of blue NOT being chosen?". In other words, "what is the probability of blue being left in the bag?"
Well that's simply 1/7. There is a 1 in 7 chance of picking blue, so there is a 1/7 chance that blue remains in the bag.
That means there is a  chance that blue is not in the bag.
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Either way, the final answer is 6/7
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