Probability – II

Here n(S) = number of elements in the sample space S = number of ways in which 2 balls can be withdrawn out of 4+6 i.e., 10 balls = ¹⁰C₂ = (10*9)/(2*1) = 45

If E be the event of drawing 2 black balls, the n(E) = the number of ways in which 2 black balls can be drawn out of 6 black balls = ⁶C₂ = (6*5)/(2*1) = 15

Required probability = n(E)/n(S) = 15/45 = 1/3

Total number of balls in the bag = 5+4+6 = 15

n(S) = number of elements in the sample space S = number of ways in which 3 balls are drawn together at random out of 15 balls = ¹⁵C₃ = (15*14*13)/(3*2*1) = 5*7*13
If E be the event of drawing one red, one white and one green ball, then n(E) = number of ways in which one red ball out of 5 red balls, one white ball out of 4 white balls and one green ball out of 6 green balls can be drawn

Total number of balls = 6+7+5 = 18

n(S) = ¹⁸C₃ = (18*17*16)/(3*2*1) = 48*17

Total number of = 6+4+8 = 18

Out of these, 3 balls can be drawn in ¹⁸C₃ ways, so the number of exhaustive cases = ¹⁸C₃ = (18*17*16)/(3*2*1) = 17*48 = n(S)

 

i. 3 red balls can be drawn in ⁶C₃ ways, so if E₁ be the event of drawing 3 red balls, then n(E₁) = ⁶C₃ = 20

P(E₁) = n(E₁)/n(S) = 20/(17*48) = 5/204

 

Let E₁ & E₂ be the events that the first card drawn and the second card drawn is an ace of heart respectively.

Since both the cards are drawn simultaneously from the same set, so both of them cannot be the ace of heart, thus E₁ and E₂ are mutually exclusive.

Also, P(E₁) = 1/52 = P(E₂), as there is only one ace of heart in the pack of 52 cards.

Required probability = P(E₁ ∪ E₂) = P(E₁)+P(E₂) = 1/52 + 1/52 = 2/52 = 1/26

Here n(S) = number of elements in the sample space S = number of ways in which one ticket can be drawn out of 10,000 = 10,000.

Let E be the event that the number marked on the drawn ticket is divisible by 20, so we have

n(E) = number of those drawn tickets, the numbers marked on which are divisible by 20 = 500 (obtained by dividing 10000 by 20)

P(E) = (n(E))/(n(S)) = 500/10,000 = 1/20

If H and T denote the appearance of ‘head’ and ‘tail’ on tossing a coin, then if we toss two coins, the possible results are 2*2 i.e. 4 and so the sample space S can be written as
S = {(H,H), (H,T), (T,H), (T,T)}
Where (H,T) means ‘head’ appearing on first coin and ‘tail’ on second.
Now, according to the problem E₁ = {(H,H), (H,T)} and E₂ = {(H,H), (T,H)}
E₁ ∪ E₂ = {(H,H), (H,T), (T,H)}
-> n(S) = 4 and n(E₁ ∪ E₂) = 3
-> P(E₁ ∪ E₂) = n(E₁ ∪ E₂)/n(S) = 3/4

i. A and B are mutually exclusive
Therefore, P(A∪B) = P(A) + P(B)
-> 0.7 = 0.4+x
-> x = 0.7-0.4 = 0.3
-> x = 3/10

 

ii. P(A∪B) = P(A) + P(B) – P(A∩B)
-> 0.7 = 0.4 +x – P(A∩B)
-> P(A∩B) = x-0.3 —– ❶
Also, if A and B are independent events, then P(A∩B) = P(A). P(B)
-> x-0.3 = (0.4) x, from ❶ and given values
-> (1-0.4) x = 0.3
-> x = 0.3/0.6 = 1/2

There are six numbers 1, 2, 3, 4, 5, 6 which can appear on the upper face of each dice, so combining the outcomes of both the dice, the total number of outcomes = 6*6 = 36 and so if S be the sample space S of the experiment, then n(S) = 36, where S = {(1,1), (1,2), ….., (1,6), (2,1), ….., (5,6), (6,1), (6,2) ….., (6,6)}, where
(5, 6) means 5 appearing on the first die and 6 appearing on the second die.

 

i. Let E₁ be the event that sum of number on the dice is 7, then
E₁ = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
Therefore, n(E₁) = 6 and so P(E₁) = n(E₁)/n(S) = 6/36 = 1/6
And E₂ be the event that sum of numbers on the dice is 8, then
E₂ = {(2,6), (3,5), (4,4), (5,3), (6,2)}
n(E₂) = 5 and so P(E₂) = n(E₂)/n(S) = 5/36
Events E₁ and E₂ are mutually exclusive, so the required probability =P(E₁ ∪ E₂) = P(E₁) + P(E₂) = 1/6 + 5/36 = 11/36

 

ii. Let E₁ be the event that the number on the first die is even, then we have
E₁ = {(2,1), (2,2) ….., (2,6), (4,1), (4,2), ….., (4,6), (6,1), (6,2), ….., (6,5), (6,6)}
n(E₁) = 18 and so P(E₂) = n(E₂)/n(S) = 18/36
Also, if E₂ be the event that the sum of numbers on the dice is 8, then E₂ = {(2,6), (3,5), (4,4), (5,3), (6,2)}
n(E₂) = 5 and so P(E₂) = n(E₂)/n(S) = 5/36
Also, E₁ and E₂ are not mutually exclusive, as they have three common elements e.g. (2,6), (4,4), (6,2)
n (E₁ ∩ E2₂) = number of elements common to E₁ and E₂ i.e. n(E₁ ∩ E₂) = n(E₁ ∩ E₂)/n(S) = 3/36
Required probability = P((E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂) = 18/36 + 5/36 – 3/36 = 20/36 = 5/9

 

iii. If E₁ and E₂be the events that the sum of numbers on the dice are 7 and 11 respectively, then
E₁ = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} and E₂ = {(5,6), (6,5)}
-> n(E₁) = 6, n(E₂) = 2
-> P(E₁) = n(E₁)/n(S) = 6/36; P(E2) = n(E₂)/n(S) = 2/36
Also, as E₁ and E₂ are mutually exclusive events, so we have
P(E₁ ∪ E₂) = P(E₁) + P(E₂) = 6/36 + 2/36 = 8/36 = 2/9
The required probability = probability of non-occurrence of both E₁ and E₂ = 1 – P(E₁ ∪ E₂) = 1 – (2/9) = 7/9

In a pack there are 52 cards out of which 13 are spades, 12 are court and 4 are deuce. So, if E₁, E₂ and E₃ be the events of a drawing a spade, a court and a deuce, and S be the sample space, then n(S) = 52, n(E₁) = 13, n(E­₂) = 12 and n(E₃) = 4

P(E₁) = n(E₁)/n(S) = 13/52

P(E₂) = n(E₂)/n(S) = 12/52

P(E₃) = n(E₃)/n(S) = 4/52

Here the events E₂ and E₃ are mutually exclusive, but the events E₁ and E₂ or E₁ and E₃ are not so.

-> n(E₂ ∩ E₃) = 0

And n(E₁ ∩ E₂) = number of elements common in E₁ and E₂ = 3, since there are 3 court cards of spade; similarly n (E₁ ∩ E₃) = 1, since there is one deuce card of spade.

Also, n (E₁ ∩ E₂ ∩ E₃) = number of elements common in E₁, E₂ and E₃ = 0, as there is no element common in three

Required probability = probability that either E₁ or E₂ or E₃ happens = P(E₁ ∪ E₂ ∪ E₃) = P(E₁) + P (E₂) + P(E₃) – P(E₁ ∩ E₂) – P(E₂ ∩ E₃) – P(E₃ ∩ E₁) + P(E₁ ∩ E₂ ∩ E₃)

= 13/36 + 12/36 + 4/36 – n(E₁ ∩ E₂)/n(S) – n(E₂ ∩ E₃)/n(S) – n(E₃ ∩ E₁)/n(S) + n(E₁ ∩ E₂ ∩ E₃)/n(S) = 13/52+ 12/52+ 4/52 – 3/52 – 0/52 – 1/52 + 0/52 = 25/52