#### DrTacoElf

10+ Year Member
15+ Year Member
You have a bag with 35 marbles, 8 of which are black.

After 5 marbles are randomly selected what is the probability that at least one black marble is drawn.

I know the answer is near 75%.

Please how is this setup #### Mr Reddly

##### Snowglobe!
7+ Year Member
15+ Year Member
I'll give it a shot ( been a loooooooooong time)

the probability of picking atleast 1 black is the same as 1 - the prob of picking all whites.
1) answer = (1 - prob of all whites)
How do we find the probability of picking white 5 times in a row?
On the first pick, what is the prob of picking a white marble?(27 white / 35 tot)
2) 27/35
On the second pick, what is the prob of a white? We already got the first white so (26 white/34 tot)
3) 26/34
And on the 3rd pick? We pulled out 2 whites already soo (25w/33tot)
4) 25/33
etc
5) 24/32
6) 23/31
Now we have the probabilities of each pick being white. Just multiply them
27/35*26/34*25/33*24/32*23/31 is the probability of picking white 5 times in a row.
Now, to find the probability of at least 1 black (ie, not getting 5 whites in a row) is
1-(27/35*26/34*25/33*24/32*23/31)

Now if that is right or not.... Haven't the slighest clue.

#### DrTacoElf

10+ Year Member
15+ Year Member
OMG, your a life saver This is a concept i've been confused about for some time kinda.

#### Mo007

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DrTacoElf said:
OMG, your a life saver This is a concept i've been confused about for some time kinda.
You have to recall one of the Probability Rules: The Multiplication Rule.

The multiplication rule states that the probability of a compound event is equal to the multiple of the probabilities of the separate parts of the event.

A compound event is a multiple-part event... (like flipping a coin twice).

By multiplying (like Mr Reddly showed), we extract the number of successes in the numerator, and the number of possible outcomes in the denominator.

Can't explain more than that... but I can give you a different example:

Let's use a deck of cards for example... and let's just say we need to find the probability of getting a "Queen" then a "Jack", so... to apply the rule above:

This probability will then be... p|queen then Jack| - p|Queen| * p|Jack|

#### sxr71

##### Senior Member
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As Mr. Reddly has shown (yes you are right), this problem is very easily done by looking at the failure situation and then subtracting that probability from one.

If you want a result where at least one ball is black, then the only scenario which fails to accommodate that is if all the balls are white.

Because we are looking at 5 independent events having all white balls is approximated by:

(27/35)*(27/35)*(27/35)*(27/35)*(27/35) or (27/35)^5

Since this is the DAT you should approximate in your head. Divide numerator and denominator by 9 (choose the highest common approximate denominator) and you will approximately have: (3/4)^5.

This can be converted mentally to 243/1024 or around 24% which gives you an approximate answer of 76% since 1-24% is 76%. The approximations are necessary if you don't have a calculator. Hopefully your answer choices aren't 74%, 75%, 76% etc.

You will never get a problem with such complex calculations on the real DAT since it is an easy test and they know you don't have a calculator.

The reason why you have to multiply when looking at successive independent events is because if you are looking for the same result on every trial you need it to be that result every time.

For example if you want all heads on five coin tosses, you need heads for the first toss. What is the probability of that? It is 0.5.

Then need heads on the second toss. The probability of that happening is also 0.5.

So you half a chance on the first toss and then half a chance on the second toss. Between the two throws you only have a quarter of a chance of getting all heads. If you took this to five successive tosses your chance of getting all heads is (0.5)^5 which is a very small number since it is pretty hard to get the same result 5 times in a row.

#### DrTacoElf

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15+ Year Member
Because we are looking at 5 independent events having all white balls is approximated by:

(27/35)*(27/35)*(27/35)*(27/35)*(27/35) or (27/35)^5

wouldn't that assume replacement?

#### sxr71

##### Senior Member
7+ Year Member
15+ Year Member
DrTacoElf said:
Because we are looking at 5 independent events having all white balls is approximated by:

(27/35)*(27/35)*(27/35)*(27/35)*(27/35) or (27/35)^5

wouldn't that assume replacement?

Yes, but in a situation where we have so many balls you can ignore that and go for an approximation. If you have to pick say 5 balls out of 7 then you should account for the missing ball because it will affect your result greatly. In the time that you have, you really are not in the position to do such complex calculations. In all honesty a real DAT question will never have this much calculation and these prep books really overdo things sometimes. If you want to see how real DAT questions are, see the DAT registration booklet. Generally you will never even have a question requiring you to do long division.

One thing that your question reminds me of is that the DAT always seems to have a question that tests whether you know how to treat problems involving replacement and those not involving replacement.

#### DrTacoElf

10+ Year Member
15+ Year Member
Thanks for the tips 