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The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)
- a)p + m + c = 19/20
- b)p + m + c = 27/20
- c)pmc = 1/10
- d)pmc = 1/4
Correct answer is option 'B,C'. Can you explain this answer?
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Verified Answer
The probabilities that a student passes in Mathematics, Physics and Ch...
According to the problem, m + p + c – mp – mc – pc + mpc = 3/4 …(1)
mp (1– c) + mc (1– p) + pc (1– m) = 2/5 or mp + mc + pc – 3mpc = 2/5 … (2)
Also mp + pc + mc – 2mpc = 1/2 … (3)
mp (1– c) + mc (1– p) + pc (1– m) = 2/5 or mp + mc + pc – 3mpc = 2/5 … (2)
Also mp + pc + mc – 2mpc = 1/2 … (3)
(2) and (3) ⇒ 
∴ mp + mc + pc = 
∴ m + p + c =
Most Upvoted Answer
The probabilities that a student passes in Mathematics, Physics and Ch...
To solve this problem, let's break it down step by step:
Step 1: Define the variables
Let m represent the probability of passing in Mathematics.
Let p represent the probability of passing in Physics.
Let c represent the probability of passing in Chemistry.
Step 2: Interpret the given information
The problem states that the student has a 75% chance of passing in at least one subject. This means that the student has a 25% chance of failing in all three subjects. So, we can write the equation:
1 - (1 - m)(1 - p)(1 - c) = 0.75
The problem also states that the student has a 50% chance of passing in at least two subjects. This can be broken down into two cases: passing in two subjects and passing in all three subjects.
Case 1: Passing in exactly two subjects
The probability of passing in exactly two subjects can be calculated by multiplying the probability of passing in two subjects and the probability of failing in the third subject. So, we can write the equation:
(1 - m)(1 - p)c = 0.40
Case 2: Passing in all three subjects
The probability of passing in all three subjects can be calculated by multiplying the probabilities of passing in each subject. So, we can write the equation:
mpc = 0.50
Step 3: Solve the equations
We now have a system of three equations with three variables. We can solve this system of equations to find the values of m, p, and c.
From the first equation, we can simplify and get:
(1 - m)(1 - p)(1 - c) = 0.25
Expanding this equation, we get:
1 - m - p - c + mp + mc + pc - mpc = 0.25
Rearranging the terms, we get:
mp + mc + pc - m - p - c = 0.75
Adding mpc to both sides, we get:
mp + mc + pc - m - p - c + mpc = 0.75 + mpc
Simplifying this equation, we get:
mpc + mp + mc + pc - m - p - c = 0.75 + mpc
Factoring out common terms, we get:
mpc + (mp - m) + (mc - c) + (pc - p) = 0.75 + mpc
Factoring out m from the second term, we get:
mpc + m(p - 1) + (mc - c) + (pc - p) = 0.75 + mpc
Similarly, factoring out p and c from the third and fourth terms, we get:
mpc + m(p - 1) + c(m - 1) + p(c - 1) = 0.75 + mpc
Rearranging the terms, we get:
mpc + m(p - 1) - mpc + c(m - 1) - p(c - 1) = 0.75
Simplifying this equation, we get:
m(p - 1) + c(m - 1) - p(c - 1) = 0.75
Expanding the terms, we get:
mp - m + cm - c - pc + p = 0.
Step 1: Define the variables
Let m represent the probability of passing in Mathematics.
Let p represent the probability of passing in Physics.
Let c represent the probability of passing in Chemistry.
Step 2: Interpret the given information
The problem states that the student has a 75% chance of passing in at least one subject. This means that the student has a 25% chance of failing in all three subjects. So, we can write the equation:
1 - (1 - m)(1 - p)(1 - c) = 0.75
The problem also states that the student has a 50% chance of passing in at least two subjects. This can be broken down into two cases: passing in two subjects and passing in all three subjects.
Case 1: Passing in exactly two subjects
The probability of passing in exactly two subjects can be calculated by multiplying the probability of passing in two subjects and the probability of failing in the third subject. So, we can write the equation:
(1 - m)(1 - p)c = 0.40
Case 2: Passing in all three subjects
The probability of passing in all three subjects can be calculated by multiplying the probabilities of passing in each subject. So, we can write the equation:
mpc = 0.50
Step 3: Solve the equations
We now have a system of three equations with three variables. We can solve this system of equations to find the values of m, p, and c.
From the first equation, we can simplify and get:
(1 - m)(1 - p)(1 - c) = 0.25
Expanding this equation, we get:
1 - m - p - c + mp + mc + pc - mpc = 0.25
Rearranging the terms, we get:
mp + mc + pc - m - p - c = 0.75
Adding mpc to both sides, we get:
mp + mc + pc - m - p - c + mpc = 0.75 + mpc
Simplifying this equation, we get:
mpc + mp + mc + pc - m - p - c = 0.75 + mpc
Factoring out common terms, we get:
mpc + (mp - m) + (mc - c) + (pc - p) = 0.75 + mpc
Factoring out m from the second term, we get:
mpc + m(p - 1) + (mc - c) + (pc - p) = 0.75 + mpc
Similarly, factoring out p and c from the third and fourth terms, we get:
mpc + m(p - 1) + c(m - 1) + p(c - 1) = 0.75 + mpc
Rearranging the terms, we get:
mpc + m(p - 1) - mpc + c(m - 1) - p(c - 1) = 0.75
Simplifying this equation, we get:
m(p - 1) + c(m - 1) - p(c - 1) = 0.75
Expanding the terms, we get:
mp - m + cm - c - pc + p = 0.
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The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)a)p + m + c = 19/20b)p + m + c = 27/20c)pmc = 1/10d)pmc = 1/4Correct answer is option 'B,C'. Can you explain this answer?
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The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)a)p + m + c = 19/20b)p + m + c = 27/20c)pmc = 1/10d)pmc = 1/4Correct answer is option 'B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)a)p + m + c = 19/20b)p + m + c = 27/20c)pmc = 1/10d)pmc = 1/4Correct answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)a)p + m + c = 19/20b)p + m + c = 27/20c)pmc = 1/10d)pmc = 1/4Correct answer is option 'B,C'. Can you explain this answer?.
The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)a)p + m + c = 19/20b)p + m + c = 27/20c)pmc = 1/10d)pmc = 1/4Correct answer is option 'B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)a)p + m + c = 19/20b)p + m + c = 27/20c)pmc = 1/10d)pmc = 1/4Correct answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)a)p + m + c = 19/20b)p + m + c = 27/20c)pmc = 1/10d)pmc = 1/4Correct answer is option 'B,C'. Can you explain this answer?.
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