A bag contains 4 red marbles and 2 blue marbles.
Blue andred marbles bag problem.
If the first marble drawn was a red marble what is the chance that the second draw is a blue marble.
This is called probability without replacement or dependent probability.
So they say the probability i ll just say p for probability.
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.
2 making a table 6 rp 3a 6 ee 7 we are given that for every three blue marbles in the bag there are two red marbles.
You reach into the bag and draw a marble and then draw another marble without replacing the first one.
How many blue marbles are there.
So simple multiplication will give the desired probability.
How many red marbles are there in the bag.
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Answer by fombitz 32378 show source.
The number of blue marbles is 1 less than 3 times the number of red marbles.
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.
The sample space for the second event is then 19 marbles instead of 20 marbles.
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What fraction of the marbles in the bag are blue.
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.
The probability of picking a yellow marble.
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.
Initially there were the same number of blue marbles and red marbles in a bag.
Let x the number of draws.
So frac 4 7 of the marbles are now red.
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John took out 5 blue marbles and then there were twice as many red marbles as blue marbles in the bag.
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.
A bag has 3 blue marbles and 4 red marbles.
Initially blue marbles red marbles x then john.
Let x red marbles.
A bag has 16 blue 20 red and 24 green marbles.
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