The probability that a student passes the Probability and St

veksetz

veksetz

Answered question

2021-12-17

The probability that a student passes the Probability and Statistics exam is 0.7.
(i)Find the probability that a given student will pass the Probability and Statistics exam after the third try.
(ii)Find the probability that the third student who fails the Probability and Statistics exam is found after the fifth student in a class of 9 students?
(iii)Find the probability that at most 3 students will pass the Probability and Statistics exam in a class of 7 students.

Answer & Explanation

Ronnie Schechter

Ronnie Schechter

Beginner2021-12-18Added 27 answers

Step 1
The probability of passing the exam is p=0.7, so the probability of failing the exam will be complement to p=0.7. Thus the probability of failing the exam can be found by subtracting p=0.7 from 1.
q=1p
=10.7
=0.3
Now, it is asked to find the probability that the student passes after third try, means the students fails in two trials but passes in third. So the probability will be equal to multiplication of fail in starting two trails and probability of pass in third try.
P=(First try)(Second try)(Third try)
=qqp
=0.30.30.7
=0.063
Step 2
Now, third student who failed in exam is found after fifth student that means there are 2 students among starting 5 students who failed the exam and 1 student in the remaining 4 students who failed the exam.
To find the probability that 2 students among starting 5 students failed the exam, apply the binomial distortion formula and similarly for probability of 1 student in the remaining 4 students failed the exam. Then multiply both the probabilities to find the probability that third student found after fifth student.
P=(52)(0.3)2(0.7)52×(41)(0.3)1(0.7)41
=5!2!(52)!0.090.73×4!1!(41)!0.30.73
=0.30870.4116
0.127
Step 3
Now, to find the probability that maximum 3 students pass the exam among 7 students, add the probability of 0, 1, 2 and 3 students pass the exam.
To find the probability that x students pass the exam among 7 students apply the binomial distribution formula.
P(X3)=P(0)+P(1)+P(2)+P(3)
=7!0!(70)!(0.7)0(0.3)70+7!1!(71)!(0.7)1(0.3)71+7!2!(72)!(0.7)2(0.3)72+7!3!(73)!(0.7)3(0.3)73
=110.37+76!16!0.70.36+765!25!0.490.35+7654!64!0.3430.34
=0.0002187+0.0035721+0.0250047+0.096957
0.126
Matthew Rodriguez

Matthew Rodriguez

Beginner2021-12-19Added 32 answers

Step 1
It is given that the probability of passing the exam is p=0.7 It is asked to find the probability that the student does not pass the exam in first two try but passed in third try.
So the Probability will be equal to multiplication of probability of fail in first try, fail in second try and pass in third try. The probability of failing the exam is complement to probability of passing the exam. So the probability fo failing the exam will be 1p
P(Passing exam in third try)=(10.7)×(10.7)×0.7
=0.3×0.3×0.3
=0.063
Step 2
For second part, 3 students among 9 could not pass the exam. It is asked to find the probability that 2 students are in first 5 students and 1 students is in remaining 4 students.
Binomial probability distribution can be used to find the probability of 2 students are in first 5 students, using total trials n=5, number of success x=2 and probability of success p=0.3.
Similarly, to find the probability that 1 students is in remaining 4 students can be found using binomial distribution by total trials n=4, number of success x=1 and probability of success p=0.3. And multiply both the probabilities to find the probability that the third student who failed exam found after fifth student.
P=5!2!(52)!(0.3)2(0.7)52×4!1!(41)!(0.3)1(0.7)41
=0.3087×0.4116
0.127
Step 3
To find the probability that maximum 3 students pass the exam among 7 students, add the probability that 0, 1, 2 and 3 students pass the exam. To find the probability that x students pass the exam out of 7 students use binomial distribution formula for 0.7 probability of success.
P(X3)=P(0)+P(1)+P(2)+P(3)
=7!0!(70)!(0.7)0(0.3)7+7!1!(71)!(0.7)1(0.3)6+7!2!(72)!(0.7)2(0.3)5
+7!3!(73)!(0.7)3(0.3)4
=1×(0.3)7+7×0.7×(0.3)6+21×0.72×(0.3)355×0.73×(0.3)4
0.126

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