True or False and how do you find the answer.

Question

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.

Answer 1

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.

 

Answer 2

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 0.9¹⁰

 

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 0.9¹⁰ + 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 =  _nC_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 = ₁₀C₁(0.9)⁹(0.1)¹

0 incorrect verdicts = ₁₀C₀(0.9)¹⁰(0.1)⁰

 

P(y<2) = P(y = 1) + P(y=0) = ₁₀C₁(0.9)⁹(0.1)¹ + ₁₀C₀(0.9)¹⁰(0.1)⁰ = 0.736

Answer 3

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 = ₁₀C₉ = 10!/((1! * 9!)) = 10

 

 

.3487 

.3874

 

.7361

 

Round to .736.

Answer 4

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.

 

Answer 5

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) 0.1⁰ 0.9¹⁰ + C(10,1) 0.1¹ 0.9⁹ = 0.9¹⁰ + 0.9⁹ = .736, so TRUE