If A and B are two events of sample space S, then
If E, F and G are events with P(G) ≠≠ 0 then P ((E ∪ F)G) given by
If A, B and C are three events of sample space S, then
If A and B are two events such that P(A) ≠ O and P(A) ≠ 1, then
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')
Let E and F be events of a sample space S of an experiment, then P(E’/F) = …
A fair sixsided die is rolled twice. What is the probability of getting 2 on the first roll and not getting 4 on the second roll?
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
In a box containing 100 Ipods, 10 are defective. The probability that out of a sample of 5 Ipods, exactly 1 is defective is:
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}
Three coins are tossed. If at least two coins show head, the probability of getting one tail is:
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
If A and B are two events such that P(A) = 0.3 and P(B) = 0.9 and P(BA) = 0.6,then P(AB) = ……
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
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:
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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