D R. D R E W M A T H 2 P R O J E C T: P R O G R A M M I N G A N D P R O B A B I L I T Y
Colorblind: SOLUTIONS
High Tech High North County has 481 students (274 boys and 207 girls). A Grade 11 biology student was
investigating the probability of a student being (at least partially) colorblind. She surveyed every student at the
school, and her data is summarized in the Two-Way Table below:
Two-Way Table
Is the
student
colorblind?
Student Gender
Total
Male
Female
Yes
39
1
40
No
235
206
441
274
207
481
Total
Using this table answer the following questions:
1. What is the probability that a student is colorblind? Not colorblind?
Pr[C] = 40 / 481 = 8.316 % Pr[NC] = 441 / 481 = 91.68 %
2. What is the probability a randomly chosen student at HTHNC is male? Female?
Pr[M] = 274 / 481 = 56.96 % Pr[F] = 207 / 481 = 43.04 %
3. What is the probability that a male student is colorblind? Not colorblind?
Pr[C M] = 39 / 274 = 14.23 % Pr[NC M] = 235 / 274 = 85.77 %
4. What is the probability that a female student is colorblind? Not colorblind?
Pr[C F] = 1 / 207 = 0.4831 % Pr[NC F] = 206 / 207 = 99.52 %
5. What is the probability that a colorblind student is male? Female?
Pr[M C] = 39 / 40 = 97.50 % Pr[F C] = 1 / 40 = 2.50 %
6. What is the probability that a student who is not colorblind is male? Female?
Pr[M NC] = 235 / 441 = 53.29 % Pr[F NC] = 206 / 441 = 46.71 %
7. What is the probability that a student is male and colorblind?
Pr[M and C] = 39 / 481 = 8.108
8. What is the probability that a student is male and not colorblind?
Pr[M and NC] = 235 / 481 = 48.86 %
9. What is the probability that a student is female and colorblind?
Pr[F and C] = 1 / 481 = 0.2079 %
10. What is the probability that a student is female and not colorblind?
Pr[F and NC] = 206 / 481 = 42.83 %
Summarize your probability expressions using probability notation and the following abbreviations:
◼ M = Male and F = Female
◼ C = Colorblind and NC = Not colorblind
For example, your answers should be in the form P(M ) = …, P(C F) = … and so on.
Challenge
1. What is the probability a student is male or colorblind? Male or not colorblind?
Pr[M or C] = Pr[M ] + P[C] - P[M and C] =
274
481
+ 40 - 39 = 275 = 57.17 %
481
Pr[M or NC] = Pr[M ] + P[NC] - P[M and NC] =
274
481
481
481
+ 441 - 235 = 480 = 99.79 %
481
481
481
2. What is the probability a student is female or colorblind? Female or not colorblind?
Pr[F or C] = Pr[F] + P[C] - P[F and C] =
207
481
+ 40 - 1 = 246 = 51.14 %
Pr[F or NC] = Pr[F] + P[NC] - P[F and NC] =
481
207
481
481
481
+ 441 - 206 = 442 = 91.89 %
481
481
481
3. Our team has 65 students (41 males and 24 females). Based on the results in the table, what is the probability
at least one student is colorblind?
Pr[At least one C] = 1 - Pr[no males and no females are colorblind ]
= 1 - Pr[no colorblind males] · Pr[no colorblind females]
= 1 - (Pr[NC M )41 (Pr[NC F)24
235 41 206 24
= 1274
207
= 99.836 %