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In a bag, there are 6 white disks, 6 black disks and 8 red disks. A disk is drawn at random from the bag.The colour is recorded and the disc is returned to the bag This process is repeated 10 times. Find the probability that less than 4 red disks are drawn

I approached this question with n= 8 and p=0.4, q=0.6. But I'm not sure about what to do with the repetition of 10 times, how do I do this question?

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As the disc is returned to the bag it will be the same probability of a red disc coming out each time. As it is less than four red discs, we need to find the probability of $0, 1, 2$ and $3$ red discs being chosen. Your values of $p$ and $q$ are indeed correct, however I'm a bit confused as to why you've chosen $n = 8$. There are $10$ trials, and as such I have chosen $n = 10$ and $r = 0, 1, 2$ and $3$, and putting this into the formula $$^nC_r\times p^{r}\times q^{n-r}$$ and adding up the four values for r you should get your answer.

Hope this helps :)

EDIT: Answer:

$(^{10}C_0 \times 0.4^0 \times 0.6^{10-0}) + (^{10}C_1 \times 0.4^1 \times 0.6^{10-1})...$
I work out the answer to be 0.382...

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  • $\begingroup$ thanks, I tried this method first but I made some calculation error which then led me to the idea that I was using the wrong formula. Haha, thanks $\endgroup$ – warman Aug 7 '17 at 9:25
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The probability P(R) of drawing a red disk is 8/20, which = 0.4. You have 10 trials, therefore:

R ~ B(10, 0.4).

Using this distribution, you want to calculate:

P(R < 4), or P(R ≤ 3), since it's discrete.

Modern calculators will evaluate this, but since the formula only uses '=' rather than '≤', you are technically doing:

P(R = 0) + P(R = 1) + P(R = 2) + P(R = 3).

The formula for P(R = r), where R ~ B(n, p) is:

$$ \binom{n}{r} \times p^r \times q^{n-r}, where \binom{n}{r} = \frac{r!}{r!(n-r)!} $$

Use this formula to sum the probabilities, and you should achieve the answer:

0.3822806016, or 0.382 (3 S.F).

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