^{x}* (1 - 0.1)

^{(10 - x)}

^{0}* (0.9)

^{10}= [1] * (1) * 0.3487ish = 0.3487ish

^{1}* (0.9)

^{9}= [10] * (0.1) * 0.3874ish = 0.3874ish

Suppose that a judges decisions follow a binomial distribution and that his verdict is correct 90% of the time. In his next 10 decisions, the probability that he makes fewer than 2 incorrect verdicts is 0.736.

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The probability that he makes fewer than 2 incorrect decisions is the same as the probability that he makes 0 or 1 incorrect decision:

P(fewer than 2 incorrect) = P(0 incorrect) + P(1 incorrect)

The binomial probability that he makes x incorrect decisions out of 10, with a probability of making an incorrect decision of 10% is:

P(x) = [(10!) / (x! * (10 - x)!)] * (0.1)^{x} * (1 - 0.1)^{(10 - x)}

P(0) = [(10!) / (0! * 10!)] * (0.1)^{0} * (0.9)^{10} = [1] * (1) * 0.3487ish = 0.3487ish

P(1) = [(10!) / (1! * 9!)] * (0.1)^{1} * (0.9)^{9} = [10] * (0.1) * 0.3874ish = 0.3874ish

P(0) + P(1) = 0.3487 + 0.3874 = 0.7360ish

So, true.

Binomial distribution uses a p and q to denote the probability of something happening and the probability that something doesn't happen. If he is correct 90% o the time then then probability that he is correct is 0.9 and the probability that he is incorrect is 0.1 or 10% of the time. 0.9 +0.1 = 1.

So we say that p = 0.9 and q = 0.1.

Since there are 10 decisions and we are looking for fewer than 2 incorrect verdicts, that mean we want the probability that he has 1 incorrect verdict or 0 incorrect verdicts.

If he has 0 incorrect verdicts then he has all correct verdicts which is (p)(p)(p)(p)(p)(p)(p)(p)(p)(p) or 0.9^{10}

If he has 1 incorrect verdict then it would be something like (q)(p)(p)(p)(p)(p)(p)(p)(p)(p) or (0.1)^{1}(0.9)^{9} An additional consideration we must make is that we don't know when the judge will make this incorrect verdict it could be (p)(p)(p)(p)(q)(p)(p)(p)(p)(p). Because of this we must multiply the (0.1)^{1}(0.9)^{9} by
**10** since there are 10 trials the
q could end up.

Then we add our results 0.9^{10} + **10**(0.1)^{1}(0.9)^{9} =0.736 , therefore the answer is True

Short-hand way of explaining:

Binomial distribution for P(y = k) for *n* trials = _{n}C_{k}(p)^{n}(q)^{n-k}, *C means combination (if you are not sure what a combination is, that is another topic probability)

Fewer than 2 means 1 or 0.

1 incorrect verdict = _{10}C_{1}(0.9)^{9}(0.1)^{1}

0 incorrect verdicts = _{10}C_{0}(0.9)^{10}(0.1)^{0}

P(y<2) = P(y = 1) + P(y=0) = _{10}C_{1}(0.9)^{9}(0.1)^{1} +
_{10}C_{0}(0.9)^{10}(0.1)^{0} = 0.736

P < 2 incorrect = P of 9 or 10 correct = P(10) + P(9)

P(10) = **1**(.9)^{10 }(.1)^{0}= .3487

Note that the 1 in the P(10) equation = 10C10 = 10!/((0! * 10!)) = 1

P(9) = **10**(.9)^{9}(.1)^{1 }= .3874

Note that the 10 in the P(9) equation = _{10}C_{9 }= 10!/((1! * 9!)) = 10

.3487

.3874

.7361

Round to .736.

Hi Diane! At face value (so to speak), you might think the probability is 100%. If the judge is correct 9 times out of 10 (which is 90%), then in 10 more decisions he's likely to make 1 mistake, right? However, the possibility also exists that he could make 0 mistakes, so you're actually looking for the odds that he errs 0 or 1 times out of 10.

As you may know, the binomial distribution yields the probability of the number of successes in a sequence of independent yes/no observations. You're looking for failures (incorrect verdicts), but that's just the number of court decisions (10) minus the number of successes. The formula for the probability of successes is:

P = n! p^{k}(1-p)^{(n-k)}

k!(n-k)!

where n = the total number of observations, k = number of successes, and p = probability. In your example, n = 10, k = 9 or 10, and p = 90%. In other words, you're looking for the probability of 9 or 10 correct decisions ("fewer than 2 incorrect verdicts") in the next 10 decisions, when the chance of a correct decision is 90%.

There are a couple of ways that you can solve for the probability. One approach is to calculate the probability of 9 successes, then calculate the probability of 10 successes, and then add those two probabilities together to see if the sum matches 0.736. You'll need to plug-in your values into the above formula, using (for 9 successes) 10, 9, and 0.9 for the variables n, k, and p respectively, and then using (for 10 successes) 10, 10, and 0.9. Remember that ! means factorial, so 10! will be 10*9*8*7*6*5*4*3*2*1.

An alternative is to use Excel to build a table that looks like this:

k P

0

1

2

3

4

5

6

7

8

9

10

where you fill-in the values for P using the formula shown above. For example, if 0 (in the above table) is in cell A2, then an appropriate formula would be

=(FACT(10) / (FACT(A2)*FACT(10-A2))) * 0.9^A2 * (1-0.9)^(10-A2)

since n = 10, p = 0.9, and k is in the table (starting with cell A2). Remember to sum the probabilities for 9 successes and for 10 successes, which you'll compare to the given answer of 0.736.

Hope that helps!

Bill W.

The probability that he is correct on any one verdict is 0.9. Therefore, the probability of an incorrect verdict is 0.1. With 10 decisions, you use the binomial distribution B(10,0.1;k) with probabilities

P(k) = C(10,k) 0.1^{k} 0.9^{10-k}.

Fewer than 2 incorrect verdicts means 0 or 1 incorrect verdicts, so we want

P(0) + P(1) = C(10,0) 0.1^{0} 0.9^{10} + C(10,1) 0.1^{1} 0.9^{9} = 0.9^{10} + 0.9^{9} = .736, so TRUE

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