What are chances of these outcomes happening more than once?

Question

Each time, there is a 1/8 chance of picking a red ball, a 1/16 chance of picking a blue ball and 1/32 for a green ball.

The ball picking is independent and takes place 17 times.

What are the chances of picking a red ball at least 2 times?

Same question for blue and green.

Thank you for working this out for me.

Answer 1

I'll deduce a general formula that works for all three balls, and then plug in the appropriate probabilities.

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Suppose a given color has probability p of being picked each of the 17 times. Let X be the discrete random variable counting the number of appearances of this color. Then X is a binomial random variable with n=17 and probability of success p. Therefore, the probability that X equals k is given by:

(*) Pr(X=k) = (17 choose k) p^k (1-p)^17-k

for k = 0, 1, ..., 17, where (17 choose k) is the binomial coefficient 17! / k! (17-k)!

Now, the probability that X is equal to at least two is the same as one minus the probability that X is equal to zero or one. In symbols:

Pr(X≥2) = 1 - Pr(X=0) - Pr(X=1)

Using formula (*), we obtain Pr(X=0) = (1-p)¹⁷ and Pr(X=1) = 17p(1-p)¹⁶. Plugging in these values and simplifying, we have:

Pr(X>=2) = 1 - (1-p)¹⁷ - 17p(1-p)¹⁶ = 1 - (1 + 16p)(1-p)¹⁶

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With the general formula in hand, we plug in p=1/8 for red, p=1/16 for blue, and p=1/32 for green, and obtain:

Red : Pr(X>=2) = 1 - 3*(7/8)¹⁶ ≈ 0.646

Blue : Pr(X>=2) = 1 - 2*(15/16)¹⁶ ≈ 0.288

Green : Pr(X>=2) = 1 - (3/2)*(31/32)¹⁶ ≈ 0.098

Hope this helps and let me know if you'd like more clarification on any step!

Answer 2

Let's look at the red example. If you understand this then you can apply it to the blue and green cases. This is a binomial distribution with n = 17 and p = 1/8

You could do this: P(X = 2) + P(X = 3) + ..... + P(X = 17), but there is a better way!

The probability of at least 2 red is the same as 1 - the complement. In probability notation:

P(X ≥ 2) = 1 - P(X = 0) - P(X = 1). In English it is the probability of all possibilities except none and one.

P(X = 0) = (1-p)ⁿ = (7/8)¹⁷ - all 17 are not red

P(X = 1) = _nC₁ (p)¹ (1-p)^n-1 = 17(1/8)(7/8)¹⁶ - One red and the others not red. Notice that there are 17 places where the red could be.

So, P(X ≥ 2) = 1 - (7/8)¹⁷ - 17(1/8)(7/8)¹⁶ ≈ .6458

Try apply this exact same logic to the blue and green cases. Post your work here if you like and we can check it. Good luck!

Answer 3

P(at least 2 red out of 17 times)=1-P(no red out of 17 times)-P(1 red out of 17 times)

P(no red in 17 times)=(1-1/8)^17=(7/8)^17

P(1 red out of 17 times)=(1-1/8)^16*1/8=(7/8)^16*(1/8)

P(at least 2 red out of 17 times)=1-(7/8)^17-(7/8)^16*(1/8)

for the same reasoning

P(at least 2 blue out of 17 times)=1-(15/16)^17-(15/16)^16*(1/16)

P(at least 2 green out of 17 times)=1-(31/32)^17-(31/32)^16*(1/32)