Quiz-1

Let event A = Sam wins and event B = Joan wins. Then,
P(A) = 0.59
Since if Sam wins, Joan cannot win and if Joan wins, Sam cannot win, so we can say that the events A and B are mutually exclusive. Other than these two events, there are
no any other possible outcomes. So,
P(A)+P(B) = 1
0.59+P(B) = 1
P(B) = 1-0.59 = 0.41

These two events cannot be disjoint because P(A)+P(B) >1.
P(A?B) = P(A)+P(B)-P(A?B).
An event is disjoint if P(A?B) = 0. If A and B are disjoint P(A?B) = 0.6+0.7 = 1.3
And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

Let b stand for boy and g for girl. The sample space of the experiment is
S = {(g, g), (g, b), (b, g), (b, b)}
Let E and F denote the following events :
E : ‘both the children are girls’
F : ‘at least one of the child is a girl’
Then E = {(g,g)} and F = {(g,g), (g,b), (b,g)}
Now E n F = {(g,g)}
Thus P(F) = 3/4
and P (E n F )= 1/4
Therefore P(E|F) = P(E n F)/P(F) = (1/4)/(3/4) = 1/3

P(getting a prize) = 10 / (10+25)
‹=› 10 / 35
‹=› 2 / 7.

Total no. of Digits = 12. Equally likely cases = 12.
There are six odd digits. Probability = 6 / 12 = 1 / 2

The probability that Soumya will get marry than Alma will be =a(1-b). Similarly, The probability that Alma will get marry than Soumya will be b(1-a). The probability that
either of these two get marry =a(1-b) +b(1-a) =a+b-2ab.

Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = n(E)/n(S) = 9/20.

The probability that first ball is white:
=12C1/30C1
=1230
=25
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:
=18C1/29C1
=18/29
Required probability,
=(25)×(18/29)
=36/145

We have 1,8,27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is 4/100 = 1/25

Total number of cases = 6*6 = 36
Favoured cases = [(3,6), (4,5), (6,3), (5,4)] = 4
So probability = 4/36 = 1/9

Let event A = Sam wins and event B = Joan wins. Then,
P(A) = 0.59
Since if Sam wins, Joan cannot win and if Joan wins, Sam cannot win, so we can say that the events A and B are mutually exclusive. Other than these two events, there are
no any other possible outcomes. So,
P(A)+P(B) = 1
0.59+P(B) = 1
P(B) = 1-0.59 = 0.41

These two events cannot be disjoint because P(A)+P(B) >1.
P(A?B) = P(A)+P(B)-P(A?B).
An event is disjoint if P(A?B) = 0. If A and B are disjoint P(A?B) = 0.6+0.7 = 1.3
And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

Let b stand for boy and g for girl. The sample space of the experiment is
S = {(g, g), (g, b), (b, g), (b, b)}
Let E and F denote the following events :
E : ‘both the children are girls’
F : ‘at least one of the child is a girl’
Then E = {(g,g)} and F = {(g,g), (g,b), (b,g)}
Now E n F = {(g,g)}
Thus P(F) = 3/4
and P (E n F )= 1/4
Therefore P(E|F) = P(E n F)/P(F) = (1/4)/(3/4) = 1/3

P(getting a prize) = 10 / (10+25)
‹=› 10 / 35
‹=› 2 / 7.

Total no. of Digits = 12. Equally likely cases = 12.
There are six odd digits. Probability = 6 / 12 = 1 / 2

The probability that Soumya will get marry than Alma will be =a(1-b). Similarly, The probability that Alma will get marry than Soumya will be b(1-a). The probability that
either of these two get marry =a(1-b) +b(1-a) =a+b-2ab.

Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = n(E)/n(S) = 9/20.

The probability that first ball is white:
=12C1/30C1
=1230
=25
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:
=18C1/29C1
=18/29
Required probability,
=(25)×(18/29)
=36/145

We have 1,8,27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is 4/100 = 1/25

Total number of cases = 6*6 = 36
Favoured cases = [(3,6), (4,5), (6,3), (5,4)] = 4
So probability = 4/36 = 1/9