SOLUTION: A bag contains 5 yellow marbles, 16 green marbles, and 2 red marbles. What is the chance of drawing a red marble? If a red marble is drawn the first time and then a second marble i
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-> SOLUTION: A bag contains 5 yellow marbles, 16 green marbles, and 2 red marbles. What is the chance of drawing a red marble? If a red marble is drawn the first time and then a second marble i
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Question 268127: A bag contains 5 yellow marbles, 16 green marbles, and 2 red marbles. What is the chance of drawing a red marble? If a red marble is drawn the first time and then a second marble is drawn without replacement, what is the probability of drawing a green marble?
You can put this solution on YOUR website! You actually have to imagine doing it and
thinking about what's in the bag at each step.
There are marbles in
the bag to start with.
You don't know what you're getting when you pull
one out. Suppose all the marbles were red?
Probability = (what's available from choices)/(total of choices available)
What is the probability now of NOT getting a red
marble?
P = 0 reds / 23 marbles
P = 0
What is the probability of getting a red one?
P = 23 reds/23 reds
P = 1
This is always true:
P = 0 when there is no chance you'll get what you want
P = 1 when you are sure to get what you want
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In your problem:
You have 1 opportunity to draw a red marble
Right away, you know it's more likely it won't be red,
since 21 aren't red and 2 are red
From above,
Probability = (what's available from choices)/(total of choices available)
P = 2 reds / (5 yellow + 16 green + 2 red) Probability of drawing a red with 1 draw
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Now I'm going to draw again. I've aready got the red, so
I don't have to recalculate it's P.
How many marbles are left?
How many of the left are green?
There are green
The P of getting a green one on the 2nd draw is
