Question

In the final exams, 40% of the students failed chemistry, 25% failed physics, and 19% failed both chemistry and physics. What is the probability that a randomly selected student failed physics given that he passed chemistry?
I have answered the question in the image below, but I would like to know if it is correct. If it is not, please include an explanation of why, as well as the step by step to get the correct answer

Answer 1

Answer:

  10%

Step-by-step explanation:

If 40% failed chemistry then 60% passed chemistry.

If 19% failed both chemistry and physics, and 25% failed physics, then 6% passed chemistry and failed physics.

If we let p' represent failing physics and c represent passing chemistry, then ...

  P(p'|c) = P(p'c)/P(c)

  P(p'|c) = 6%/60% = 0.10 = 10%

If the randomly chosen student passed chemistry, the probability is 10% that he failed physics.

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Your answer is correct.

Answer 2

Answer:

I did a similar question for my homework, you can check mine.

Step-by-step explanation:

In an examination 30% of the students failed in Physics, 25% in Mathematics and 12% in both Physics and Mathematics. A student is selected at random. Find the probability that the student has failed in Physics, if it is known that he has failed in Mathematics. (ii) the student has failed at least one of the two subjects (ii) the student has passed at least one of the two subjects (iv) the student has passed in Mathematics if he failed in Physics.

Ans:

let the event of student failed in physics be P

let the event of student failed in maths be M

P(P) = 3/10

P(M)= 1/4

P(P nn M) =3/25

(I)

P(P/M)= (P(P nn M))/(P(M)) = (3/25)/(1/4) = 12/25

(ii)

P(P uu M) = P(P) + P(M) -P(P nn M)

= 3/10 + 1/4 - 3/25 = 43/100 = 0.43

(iii)

P=1 - P(P nn M) = 1 - 12/100 = 88/100 =22/25

(iv)

p-> 30 , m-> 25

P nn M -> 12

P= 18/30 = 3/5