QUESTION IMAGE
Question
giving a test to a group of students, the grades and gender are summarized below. round your answers to 4 decimal places.
grades and gender
| a | b | c | total | |
|---|---|---|---|---|
| female | 14 | 20 | 15 | 49 |
| total | 18 | 36 | 21 | 75 |
if one student is chosen at random,
a. find the probability that the student got an a:
b. find the probability that the student was female and got a \c\:
c. find the probability that the student was female or got an \a\:
d. if one student is chosen at random, find the probability that the student got an b given they are female:
Step1: Recall probability formula
$P(X)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
Step2: Calculate probability for part a
The number of students who got an A is 18, and the total number of students is 75. So $P(\text{got an A})=\frac{18}{75}= 0.24$
Step3: Calculate probability for part b
The number of female students who got a C is 15, and the total number of students is 75. So $P(\text{female and got a C})=\frac{15}{75}=0.2$
Step4: Calculate probability for part c
Use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Let $A$ be the event of being female and $B$ be the event of getting an A. $P(A)=\frac{49}{75}$, $P(B)=\frac{18}{75}$, $P(A\cap B)=\frac{14}{75}$. Then $P(A\cup B)=\frac{49 + 18- 14}{75}=\frac{53}{75}\approx0.7067$
Step5: Calculate probability for part d
Use the formula for conditional - probability $P(X|Y)=\frac{P(X\cap Y)}{P(Y)}$. Let $X$ be the event of getting a B and $Y$ be the event of being female. $P(X\cap Y)=\frac{20}{75}$, $P(Y)=\frac{49}{75}$. Then $P(X|Y)=\frac{\frac{20}{75}}{\frac{49}{75}}=\frac{20}{49}\approx0.4082$
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a. $0.24$
b. $0.2$
c. $0.7067$
d. $0.4082$