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Sagot :
To find the probability that an ACME College student who takes Precalculus will pass both Precalculus and Calculus, we need to consider the two probabilities given in the problem:
1. The probability of passing Precalculus (P(Pass Precalculus)) is 41%, or 0.41 in decimal form.
2. The probability of passing Calculus given that the student has passed Precalculus (P(Pass Calculus | Pass Precalculus)) is 48%, or 0.48 in decimal form.
To find the overall probability that a student will pass both courses, we multiply these two probabilities. This is because the events are dependent (the probability of passing Calculus depends on the student having passed Precalculus), and we use the multiplication rule for dependent events.
Mathematically, we express this as:
[tex]\[ P(\text{Pass Precalculus and Pass Calculus}) = P(\text{Pass Precalculus}) \times P(\text{Pass Calculus | Pass Precalculus}) \][/tex]
Substituting in the given probabilities:
[tex]\[ P(\text{Pass Precalculus and Pass Calculus}) = 0.41 \times 0.48 \][/tex]
Performing the multiplication:
[tex]\[ 0.41 \times 0.48 = 0.1968 \][/tex]
To match the required format, we round this result to three decimal places:
[tex]\[ 0.1968 \approx 0.197 \][/tex]
So, the probability that a student at ACME College who takes Precalculus will pass both Precalculus and Calculus is [tex]\( 0.197 \)[/tex].
1. The probability of passing Precalculus (P(Pass Precalculus)) is 41%, or 0.41 in decimal form.
2. The probability of passing Calculus given that the student has passed Precalculus (P(Pass Calculus | Pass Precalculus)) is 48%, or 0.48 in decimal form.
To find the overall probability that a student will pass both courses, we multiply these two probabilities. This is because the events are dependent (the probability of passing Calculus depends on the student having passed Precalculus), and we use the multiplication rule for dependent events.
Mathematically, we express this as:
[tex]\[ P(\text{Pass Precalculus and Pass Calculus}) = P(\text{Pass Precalculus}) \times P(\text{Pass Calculus | Pass Precalculus}) \][/tex]
Substituting in the given probabilities:
[tex]\[ P(\text{Pass Precalculus and Pass Calculus}) = 0.41 \times 0.48 \][/tex]
Performing the multiplication:
[tex]\[ 0.41 \times 0.48 = 0.1968 \][/tex]
To match the required format, we round this result to three decimal places:
[tex]\[ 0.1968 \approx 0.197 \][/tex]
So, the probability that a student at ACME College who takes Precalculus will pass both Precalculus and Calculus is [tex]\( 0.197 \)[/tex].
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