Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics, and in at least one of the subjects respectively. An examinee is selected at random. Find the probabilities that
a. He failed in Mathematics only
b. He passed in Statistics, if it is known that he has failed in Mathematics
as u hve explained everything but i didn't get that 45% failed in atleast 1 of the subject but if we add
(5+25+10)% then we get only 40% which should be 45%.....plz explain
To find the probabilities in this scenario, we will use conditional probability. Let's denote the events as follows:
A: Failed in Statistics
B: Failed in Mathematics
C: Failed in at least one subject
We are given the following information:
P(A) = 0.30 (30% failed in Statistics)
P(B) = 0.35 (35% failed in Mathematics)
P(C) = 0.45 (45% failed in at least one subject)
a. To find the probability that an examinee has failed in Mathematics only (B only), we need to find P(B and not A).
P(B and not A) = P(B) - P(B and A)
We are not given the probability of P(B and A) directly, but we can use the formula:
P(B and A) = P(A|B) * P(B)
P(A|B) represents the probability of failing in Statistics given that the examinee has failed in Mathematics. Since the examinee has failed in Mathematics, we know that event B has occurred, so P(B|A) = 1.
Therefore, P(A|B) = P(A and B) / P(B) = 1 * P(B) / P(B) =1.
Using this information:
P(B and not A) = P(B) - P(B and A) = P(B) - P(A|B) * P(B) = 0.35 - (1 * 0.35) = 0.35 - 0.35 = 0.
Therefore, the probability of an examinee failing in Mathematics only is 0.
b. To find the probability that an examinee has passed in Statistics, given that he has failed in Mathematics, we need to find P(A|B).
P(A|B) = P(A and B) / P(B)
We already calculated P(A and B) = P(A|B) * P(B) = 1 * 0.35 = 0.35 in the previous step.
Using this information:
P(A|B) = P(A and B) / P(B) = 0.35 / 0.35 = 1.
Therefore, the probability of an examinee passing in Statistics, given that he has failed in Mathematics, is 1 or 100%.
To find the probabilities in this scenario, we will need to use the concept of conditional probability.
Let's denote the events:
F: failing in Statistics
M: failing in Mathematics
Now, given the information provided:
P(F) = 30% = 0.30
P(M) = 35% = 0.35
P(at least one subject) = 45% = 0.45
a. To find the probability that an examinee failed in Mathematics only, we need to find the probability of failing in Mathematics (M) but not failing in Statistics (F).
To do this, we can use the formula for the probability of the intersection of two events (in this case, the probability of failing in Mathematics but not in Statistics):
P(M only) = P(M ∩ ¬F)
However, we do not have the direct information for P(M ∩ ¬F). But we can find it using the formula for conditional probability:
P(M ∩ ¬F) = P(M | ¬F) * P(¬F),
where P(M | ¬F) is the probability of failing in Mathematics given that the examinee did not fail in Statistics, and P(¬F) is the probability of not failing in Statistics.
We are not given the value of P(M | ¬F), but it can be derived using the formula:
P(M | ¬F) = P(M ∩ ¬F) / P(¬F),
where P(M ∩ ¬F) can be calculated as P(at least one subject) - P(F ∪ M).
We know P(at least one subject) = 45%, and P(F ∪ M) = P(F) + P(M) - P(F ∩ M), where P(F ∩ M) is the probability of failing in both Statistics and Mathematics.
P(F ∩ M) can be calculated as P(F) + P(M) - P(at least one subject), since it represents the overlapping region of failing in both Statistics and Mathematics.
Now, we can plug in the values to calculate the probabilities:
P(F ∩ M) = P(F) + P(M) - P(at least one subject)
= 0.30 + 0.35 - 0.45
= 0.20
P(M ∩ ¬F) = P(at least one subject) - P(F ∪ M)
= 0.45 - (0.30 + 0.35 - 0.45)
= 0.25
P(M | ¬F) = P(M ∩ ¬F) / P(¬F)
= 0.25 / (1 - P(F))
= 0.25 / (1 - 0.30)
= 0.25 / 0.70
= 0.357
Finally, we can calculate P(M only) using the formula:
P(M only) = P(M ∩ ¬F)
= 0.25
Therefore, the probability that an examinee failed in Mathematics only is 0.25 or 25%.
b. To find the probability that an examinee passed in Statistics, given that they failed in Mathematics, we need to find the conditional probability P(F | M).
Using the formula for conditional probability, we have:
P(F | M) = P(F ∩ M) / P(M)
We already calculated P(F ∩ M) as 0.20, and P(M) from the given information is 0.35.
Now, we can calculate the probability:
P(F | M) = P(F ∩ M) / P(M)
= 0.20 / 0.35
≈ 0.571
Therefore, the probability that an examinee passed in Statistics, given that they failed in Mathematics, is approximately 0.571 or 57.1%.
Make a Venn diagram.
5% fail math only
25% fail both
10% fail stats only
60% pass both.
Of the 30% (of all students) who failed math, 5/6 of that group (25% of all students)failed stats also. That is a conditional probability of 83.3%. Only 16.7% of the math-failing group pass stats.