Need help with stats homework? 2 probability questions

These seem so simple, but I swear something is wrong with the multiple choices.

1. Find the probability that in 200 tosses of a fair die, we will obtain exactly 30 fives.
A) 0.1871
B) 0.0871
C) 0.0429
D) 0.0619

2. A coin is tossed 20 times. What is the probability someone correctly predicts the results of each coin toss 14 OR MORE times?
A) 0.0582
B) 0.4418

1) Binomial distribution, with n = 200, p = 1/6

P(exactly 30 5's out of 200 dice rolls)
= C(200,30) * (1/6)^30 * (5/6)^170
= 0.0641

2) Binomial distribution, with n = 20, p = 1/2

P(x) = P(exactly x correct predictions)
= C(20,x) (1/2)^x (1/2)^(20-x) = C(20,x) (1/2)^20

P(14 or more correct predictions)
= P(14) + P(15) + P(16) + P(17) + P(18) + P(19) + P(20)
= 0.0577

You're right, these answers do seem a little off.

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EDIT:

OK, I think I've figured it out. You're probably supposed to use a normal approximation to the binomial distributions. (Do any of the problems mention this?)

1.

X ~ B(200, 1/6)

Y ~ N(μ, σ²), where
μ = n*p = 200 * 1/6 = 100/3
σ² = n*p*(1−p) = 200*1/6*5/6 = 250/9
σ = 5√10/3

P(x = 30)
= P(29.5 < y < 30.5)
= P((29.5−100/3)/(5√10/3) < z < (30.5−100/3)/(5√10/3))
= P(−0.73 < z < −0.54)
= P(z < −0.54) − P(z < −0.73)
= 0.2946 − 0.2327
= 0.0619

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2.

X ~ B(20, 1/2)

Y ~ N(μ, σ²), where
μ = n*p = 20 * 1/2 = 10
σ² = n*p*(1−p) = 20*1/2*1/2 = 5
σ = √5

P(x ≥ 14)
= 1 − P(x < 14)
= 1 − P(y < 13.5)
= 1 − P(z < (13.5−10)/√5)
= 1 − P(z < 1.57)
= 1 − 0.9418
= 0.0582

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Normal approximation seems to give the results they are looking for.

1. The answer is from a normal approximation to the binomial -- not an exact computation.

μ = np = 200(1/6) = 33.33

σ = √npq = √200(1/6)(5/6) = 5.27

So find the area under the normal curve from 29.5 to 30.5 and get 0.061945
The exact Binomial answer is close. 0.064083

You were probably not using the method expected for the exercise.

For the second one, since N is fairly small, the large N approx does not apply and you should get 0.057659 approx. N = 20, k=14, p=0.5 and use 1– *cumulative probability (0-13)*.

I suspect whoever did it rounded off each part before adding and accumulated roundoff error. A common error.