Question

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 i selectedat random find the probability that
a. he failed in mathematics only
b. he passed in statistics if it is known that he has failed in mathematics

Answer 1

10% must have failed at both, so 20% failed stats only, 25% math only, 10% both and 45% neither

a. 25%

b. 0.15/0.25 = 0.6 = 60%

Answer 2

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.
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Answer 3

Cool problem.

Probability failed stat = 30%
Probability failed math = 35%
Probability failed at least one = 45%

need some notation lets say

P(S) is prob failed stat
P(M) is prob failed math

P(Failed at least one) = P(MuS)

we know

P(MuS) = P(M) + P(S) - P(MnS)

so

.45 = .35 + .30 - P(MnS)

means P(MnS) = .2

now for answer to a)

P(failed math only) = P(M) - P(MnS) = .35 - .2

while we are here:
P(failed stat only) = P(S) - P(MnS) = .3 - .2

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b) asks for conditional probability that passed stat given failed math.

something like:

P(Sc | M) = P(Sc n M) / P(M) {Sc is the compliment of S}

we know P(M) so we need P(Sc n M)

The dirty details:

{notation again P(pass stat) is P(Sc) Sc for compliment of fail stats.}

P(M) = P(M n S) + P(M n Sc) so;
.35 = .15 + P(M n Sc) drum roll please...

P(M n Sc) = .2

now answer b)

P(Sc | M) = P(Sc n M) / P(M) = .2/.35 Done!

I hope this helps...

Answer 4

As per the selection of student grps will be formed. Grp A : who passed both subjects Grp B : who passed stats only and failed maths only Grp C : who passed maths but failed stats Grp D : who failed both. So, the probabilty of that student failed in maths only is: P(B) = 1/total grps = 25%