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QUESTION: 1

If A and B are two events of sample space S, then

Solution:
The probability of occurrence of event A under the condition that event B has already occurred& P(B)â‰ 0 is called Conditional probability i.e; P(A|B)=P(A âˆ© B)/P(B). Multiply with P(B) on both sides implies P(A âˆ© B)=P(B).P(A|B). So option 'A' is correct.

QUESTION: 2

If E, F and G are events with P(G) ≠≠ 0 then P ((E ∪ F)|G) given by

Solution:

QUESTION: 3

If A, B and C are three events of sample space S, then

Solution:

QUESTION: 4

If A and B are two events such that P(A) ≠ O and P(A) ≠ 1, then

Solution:

As we know that, P(A') = 1 - P(A)

P(A'⋂B') = P(AUB)' = 1 - P(AUB)

P(A'/B' ) = P(A'⋂B' ) / P(B')

⇒ ( 1 - P(A U B )) / P(B')

QUESTION: 5

Let E and F be events of a sample space S of an experiment, then P(E’/F) = …

Solution:

QUESTION: 6

A fair six-sided die is rolled twice. What is the probability of getting 2 on the first roll and not getting 4 on the second roll?

Solution:

The two events mentioned are independent. The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.

P(getting first 2) = 1/6

P(no second 4) = 5/6

Therefore P(getting first 2 and no second 4) = 1/6* 5/6 = 5/36

QUESTION: 7

In a box containing 100 Ipods, 10 are defective. The probability that out of a sample of 5 Ipods, exactly 1 is defective is:

Solution:

r = 1, n = 5

p = 10/100 = 1/10,

q = 1 - p

q = 1 - 1/10 = 9/10

= nCr (p)^{r} (q)^{(n - r)}

Exactly one is defective = 5C1 (1/10)^{1} (9/10)^{(5 - 1)}

= 5C1 (1/10) (9/10)^{4}

= (1/2) (9/10)^{4}

QUESTION: 8

Three coins are tossed. If at least two coins show head, the probability of getting one tail is:

Solution:

Subset={HHH , HHT,HTH,HTT,THH,THT,TTH,TTT}

P(at least two head) = 4/8 = ½

Getting(one tail) = {HHT, HTH, THH}

= 3/8

Therefore, P(one tail) = (⅜) / (½)

= ¾

QUESTION: 9

If A and B are two events such that P(A) = 0.3 and P(B) = 0.9 and P(B|A) = 0.6,then P(A|B) = ……

Solution:

P(B/A) = P(A∩B)/P(A)

P(A∩B) = P(B/A))×P(A)

=0.6×0.3

=0.18

P(A∩B) = P(A/B))×P(B)

0.18 = P(A/B) * 0.9

P(A/B) = 0.2

QUESTION: 10

A bag contains 25 tickets numbered from 1 to 25. Two tickets are drawn one after another without replacement. The probability that both tickets will show even numbers is:

Solution:

There are 12 even numbers between 1 to 25

Consider the given events.

A = Even number ticket in the first draw

B = Even number ticket in the second draw

Now, P(A) = 12/25

P(B/A) = 11/24

Required probability : P(A⋂B) = P(A) * P(B/A)

= (12/25) * (11/24)

= 11/50

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