This is called probability without replacement or dependent probability.
Blue andred marbles bag problem.
A bag has 16 blue 20 red and 24 green marbles.
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The sample space for the second event is then 19 marbles instead of 20 marbles.
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.
John took out 5 blue marbles and then there were twice as many red marbles as blue marbles in the bag.
The number of blue marbles is 1 less than 3 times the number of red marbles.
A bag contains blue marbles and red marbles 71 in total.
So frac 4 7 of the marbles are now red.
Let x the number of draws.
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.
A bag contains 4 red marbles and 2 blue marbles.
You reach into the bag and draw a marble and then draw another marble without replacing the first one.
What is the chance that the first draw is a red marble.
And so this is sometimes the event in question right over here is picking the yellow marble.
You draw a marble at random without replacement until the first blue marble is drawn.
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So simple multiplication will give the desired probability.
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What fraction of the marbles in the bag are blue.
Let x red marbles.
If the first marble drawn was a red marble what is the chance that the second draw is a blue marble.
How many red marbles are there in the bag.
So they say the probability i ll just say p for probability.
Answer by fombitz 32378 show source.
Initially there were the same number of blue marbles and red marbles in a bag.
How many blue marbles are there.
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The probability of picking a yellow 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.
A if you repeated this experiment a very large number of times on average how many draws would you make before a blue marble was drawn.
Initially blue marbles red marbles x then john.