Typically, 10% of students make a D on their tests, 60% make a C on their tests, and 30% make an A. Mrs. Smith uses a random-number table to find the experimental probability that of 5 students, at least 3 will make a C. The digit 0 represents students who make a D. The digits 1, 2, 3, 4, 5, and 6 represent students who make a C. The digits 7, 8, and 9 represent students who make an A.
74510 62548 71205 07819 02478
15678 02389 01258 23789 34052
84150 27549 05689 07821 56892
16059 07894 12356 01548 23689
Find the experimental probability that of 5 students, at least 3 will make a C.
A. Start Fraction 7 over 20 End Fraction
B. Start Fraction 9 over 20 End Fraction
C. Start Fraction 11 over 20 End Fraction
D. Start Fraction 13 over 20 End Fraction
To find the probability of at least 3 students making a C, we need to add up the probabilities of exactly 3, 4, or 5 students making a C.
For exactly 3 students to make a C, we need to choose 3 out of 5 students to make a C and the other 2 to make either a D or an A. The probability of one student making a C is 0.6, so the probability of 3 students making a C is:
$P(3\text{ C's}) = \binom{5}{3}(0.6)^3(0.4)^2 = 0.288$
For exactly 4 students to make a C, we need to choose 4 out of 5 students to make a C and the other 1 to make either a D or an A. The probability of one student making a D is 0.1 and the probability of one student making an A is 0.3, so the probability of 4 students making a C is:
$P(4\text{ C's}) = \binom{5}{4}(0.6)^4(0.1)(0.3) = 0.108$
For all 5 students to make a C, the probability is simply:
$P(5\text{ C's}) = (0.6)^5 = 0.07776$
Adding up these probabilities, we get:
$P(\text{at least 3 C's}) = P(3\text{ C's}) + P(4\text{ C's}) + P(5\text{ C's}) = 0.288 + 0.108 + 0.07776 = 0.47376$
So the experimental probability of at least 3 students making a C is approximately:
$P(\text{at least 3 C's}) \approx 0.474$
Looking at the answer choices, the closest one is D. $\boxed{\text{(D) }\frac{13}{20}}$.