A bag contains 4 red marbles and 2 blue marbles.
Blue andred marbles bag problem.
The number of blue marbles is 1 less than 3 times the number of red marbles.
So frac 4 7 of the marbles are now red.
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You reach into the bag and draw a marble and then draw another marble without replacing the first one.
The sample space for the second event is then 19 marbles instead of 20 marbles.
Initially blue marbles red marbles x then john.
Let x the number of draws.
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You draw a marble at random without replacement until the first blue marble is drawn.
If the first marble drawn was a red marble what is the chance that the second draw is a blue marble.
For example a marble may be taken from a bag with 20 marbles and then a second marble is taken without replacing the first marble.
With our new ratio of 3 4 for blue marbles to red marbles this means that 4 out of every 7 marbles in the bag are red.
So they say the probability i ll just say p for probability.
How many blue marbles are there.
What fraction of the marbles in the bag are blue.
Initially there were the same number of blue marbles and red marbles in a bag.
Since the first marble is replaced before the second marble is drawn the colour of the second marble is independent of the colour of the first marble.
A bag has 3 blue marbles and 4 red marbles.
What is the chance that the first draw is a red marble.
John took out 5 blue marbles and then there were twice as many red marbles as blue marbles in the bag.
Find the probability of pulling a yellow marble from a bag with 3 yellow 2 red 2 green and 1 blue i m assuming marbles.
This is called probability without replacement or dependent probability.
Let x red marbles.
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How many red marbles are there in the bag.
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The probability of picking a yellow marble.
And so this is sometimes the event in question right over here is picking the yellow marble.
A bag has 16 blue 20 red and 24 green marbles.
So simple multiplication will give the desired probability.