Question

100 students appeared for two different examinations 60 passed the first,50 the second and 30 both the examinations.find the probability that a student selected at random failed in both the examination

Answer 1

Answer:

Probability that a student selected at random failed in both the examination  = 0.2

Explanation:

Total number of student given are 100.

Number of student passed in first examination is 60. So, probability of student passing in first examination is, P(A) = $\frac {60}{100}$ = 0.6

Number of student passed in second examination is 50. So, probability of student passing in second examination is, P(B) = $\frac{50}{100}$ = 0.5

Number of student passed in both examination is 30. So, probability of student passing in both examination is, P(A ∩ B) = $\frac{30}{100}$ = 0.3

From probability addition rule,

P(A ∪ B) = P(A) + P(B) - P(A∩B)

P(A ∪ B) = 0.6 + 0.5 - 0.3

P(A ∪ B) = 0.8

Thus, probability of student passing at least in one subject is 0.8.

Probability of student failing in both subject is given by,

${P(A U B)}'$ = 1 - P(A ∪ B)

${P(A U B)}'$ = 1 -0.8

${P(A U B)}'$ = 0.2

Thus, Probability of student failing in both subjects are 0.2.

Answer 2

Answer:

Explanation:

n(1)=60[no.of students passed in first exam]

n(2)=50[no.of students passed in second exam]

n(1∩2)=30[no.of students passed in both exams]  

[n(AUB)=n(A)+n(B)-n(A∩B)]  

no. of students passed = (60+50)-30  

=80  

no. of students failed in both exams = 100 - 80  

=20  

PROBABILITY = 20/100  

=1/5