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# need probability help fast!

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Every time I use a piece of scrap paper, I crumple it up and try to shoot it inside the recycling bin across the room. I'm pretty good at it: If I shoot 5 pieces of paper at the recycling bin, at least one of them will make it inside the recycling bin with probability $$\frac{211}{243}$$. If I shoot 6 pieces of paper at the recycling bin, what's the probability at least two of them make it inside the recycling bin?

(Assume that the probability I make any given shot is uniform across every shot. Leave your answer in exact form; there should be no decimals in your answer.)

Dec 8, 2019

#1
+106981
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It is a standard combination question.

At least 2 means  P(2) + P(3) + P(4)+P(5)+P(6)

Thie is exactly the same as   1 - [P(0)+P(1)]

P(n) = 6Cr  *  (211/243)^r  * (1- 211/243)^(6-r)

Now you have all the info you need to answer the question.

Explanation of late edit:

I have changed the letter to r becasue r is the one normally used and it was confusing for you before.

I appologize for that.

Dec 8, 2019
edited by Melody  Dec 8, 2019
#2
+1

Is n the number of balls you make?

Guest Dec 8, 2019
#3
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Also can you explain how you got the probability of P(n)?

Guest Dec 8, 2019
#4
+106981
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Yes n WAS the number of sucessful shots.

I probably should have used r because n is usually used for the total number of tries. - I have revised the original question now.

6 is your total number of tries.

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The formula for getting   r  sucesses in n independent trials its      (notice here i have used n to mean total number)

$$\color{red}\text{Prob (of r sucesses in n independent trials)}\\= \color{red}\;^nC_r\;*\;(\text{prob of a sucess)}^r\;*\text{(prob of a failure)}^{(n-r)}$$

n=6

prob of sucess = $$\frac{211}{243}$$

Prob of failure = $$1-\frac{211}{243}=\frac{32}{243}$$

So, for instance in your example

$$\text{Prob (2 sucesses in 6 independent trials)}\\= \;^6C_2\;*\;(\frac{211}{243})^2\;*\;(\frac{32}{243})^{(6-2)}$$

Melody  Dec 8, 2019
edited by Melody  Dec 8, 2019
edited by Melody  Dec 8, 2019
#5
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I worked this out a different way and got a weird answer. Could you check my work to see if it is correct?

In order to solve this problem we should first find the probability of making a ball into the bin on the first shot. It is not the probability of making one in 5 shots divided by 5 because you can make multiple shots out of 5.

We can find the probability of making a ball into the bin on the first shot by making an equation. First, we need to find the chance we don't make a ball in the bin in 5 shots. That is $$1-\frac{211}{243}=\frac{32}{243}$$ Next we can set p as P(making a ball into the bin on the first shot). This is the same probability of all the other shots because the probability is uniform across every shot. In terms of p, the chance of not making a ball in all of the shot is$$(1-p)^5$$. Now we can make an equation. We get $$(1-p)^5=\frac{32}{243}$$. We can take the 5th root of both sides to get $$(1-p)=\frac{2}{3}$$}. Solving for p we get $$p=1-\frac{2}{3}=\frac{1}{3}$$.

Next, we can use complementary counting to find the probability of getting at least 2 shots in. `We can say P(0) is the probability of making no shots. P(1) is the probability of making one shot and so on. We can use this to make an equation. The probability of making at least 2 shots is 1-P(0)-P(1).

The probability of making 0 shots is $$(\frac{2}{3})^6=\frac{64}{729}$$}. Next we have to find the probability of making 1 shot. The probability of that is $$(\frac{2}{3})^5 \cdot \frac{1}{3}=\frac{32}{729}$$. Now, we can find the probability of making at least 2 shots by plugging in the probabilities. We get $$1-\frac{64}{729}-\frac{32}{729}=\frac{633}{729}=\frac{211}{243}$$. So, the probability of making two shots in 6 tries is $$\frac{211}{243}$$}.

Guest Dec 9, 2019
#6
+106981
+1

Your answer is not correct. If you use the proper formula you will get it correct.

The prob of making exactly 2 shots in 6 tries is

$$\text{Prob (2 sucesses in 6 independent trials)}\\ = \;^6C_2\;*\;(\frac{211}{243})^2\;*\;(\frac{32}{243})^{(6-2)}\\~\\ = \;^6C_2\;*\;(\frac{211}{243})^2\;*\;(\frac{32}{243})^{4}\\~\\ =\quad 15\;* 211^2*\;32^4\;\div (243^6)$$

15*211^2*32^4/ 243^6 = 0.003401092481817478

That is a little lower than I expected But that is how probability goes.

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You can check if any mothod is correct by working out all the possible probabilities.

I mean

P(0 in) + P(1 in) + ......    +P(6 in)   and they have to add up to 1

Dec 9, 2019