Suppose you roll 100 dice. What are the odds that 25 or more of them come up " 6 "? Choose the best answer. ≈0.2143 ≈0.0125 ≈0.0357 ≈0.0408

Answer 1

The results for the number of 6's follows a binomial distribution B(100, 1/6).

The probability of at least 25 6's is 1 - (probability of 24 or fewer). This quantity can be found in a table of the cumulative binomial distribution or by using the probability functions on a graphing calculator. On a TI-83 or TI-84, use binomcdf(100, 1/6. 24).

1 - binomcdf(100, 1/6. 24) = .0217

None of the choices is close, but 0.0125 is the closest.

Answer 2

Hi Emma:

This is a common type of statistical problem. Here's how to handle them.

With a standard die, define the event "a six comes up" as a "success."

The probability of a "success" on each throw = (# of ways a 6 can show) / (1 of possible events)is =1/6

Casting 100 dice (assuming all have the same probability of a "success", namely, 1/6) is modeled by the binomial distribution: P{k; n; p} = ⁿC_k p^k (1 - p)^n-k [k = # of successes, n = # of trials, p=Pr {success}

And, by definition: ⁿC_k = n! / k! (n - k)!

Now, at least 25" simply means 25 or 26 or 27 or . . . or 100 successes (sixes showing).

Here's a useful trick: P{at least 25 successes} = 1 - Pr {24 or fewer successes}. Do you see why?

=1 - Pr {1 success or 2 or 3 or . . . or 24}

Since only one of these events can occur, these are 24 mutually exclusive events. Therefore, their individual probabilities add together.

Therefore, Prob {25 or more sixes} = 1 - Pr {1 to 24 successes}

= 1 - ∑[k = 1 to 24] ¹⁰⁰C_k (1/6)^k (5/6)^{100 - k}

= 1 - ∑[k = 1 to 24] 100!/k!(100-k)! 5^{100 - k} / 6¹⁰⁰

This is a messy sum to calculate. I leave the calculation to you!.

I think you will agree that the probability a large number of sixes (100 or 99 or 98 . . .) is extremely small.

Answer 3

The correct answer is approximately 0.021703378733936. You can get .0125 by using a normal approximation without a continuity correction and rounding your z value to decimal places. You can see how far off that is, though.

The correct answer is arrived at in a few different ways. For example, you can calculate 1 - P(x <= 24 with 100 trials and p = 1/6). In Excel, we would use 1 - BINOM.DIST(24,100,1/6,1)

For the way to get to .0125, we calculate, as 25 means the proportion is 1/4, (1/4-1/6)/sqrt(1/6*5/6/100) = 2.236, and 1 - P(z <= 2.24) = .0125.

Answer 4

Emma

This is a binomial distribution with n=100, p = 1/6 (chance of a 6), q=5/6.

Can be approximated by a normal. X ∼ N (np, npq) because npq > 5

np = 16.7, npq = 13.9

We want p (X>=25).

Z = (X-µ)/σ = (25 - 16.7) / 13.9 = 0.5971

So using a table of Z values / area under a standard normal, when Z=0.60, (closest to 0.5971) area to left of X=25 is 0.7257, so area to right = 1-0.7257 = 0.2743. So the best answer is the first one.

Mike