An unbiased coin is tossed three times. If A is the event that a head appears on each of the first two tosses , B is the event that a tail occurs on the third toss and C is the event that exactly two tails appears in the three tosses, show that :
i) Event A and B are independent
ii) Event B and C are dependent
To show that two events are independent, we need to prove that the probability of their intersection is equal to the product of their individual probabilities. Similarly, to show that two events are dependent, we need to demonstrate that the probability of their intersection is not equal to the product of their individual probabilities.
Let's start by finding the probabilities for each event:
i) Event A: Head on the first two tosses.
The probability of getting a head on each toss is 1/2, as the coin is unbiased.
So the probability of event A is (1/2) * (1/2) = 1/4.
ii) Event B: Tail on the third toss.
The probability of getting a tail on the third toss is also 1/2.
So the probability of event B is 1/2.
iii) Event C: Exactly two tails in three tosses.
To calculate the probability of event C, we need to find the number of ways two tails can occur in three tosses.
The three possible arrangements are: HHT, HTH, THH, where H represents a head and T represents a tail.
Out of these three possibilities, only HTH satisfies event B, as it has a tail on the third toss.
So the probability of event C is 1/3.
Now let's calculate the probabilities of the intersections of these events:
i) Intersection of event A and B: Head on first two tosses and tail on the third toss.
The probability of the intersection is (1/2) * (1/2) * (1/2) = 1/8.
ii) Intersection of event B and C: Tail on the third toss and exactly two tails in three tosses.
The probability of the intersection is 1/3, as there is only one possibility (HTH) that satisfies both events B and C.
Comparing the probabilities of the intersections with the product of the individual probabilities:
i) (1/8) ≠ (1/4) * (1/2)
Therefore, event A and B are not independent.
ii) (1/3) = (1/2) * (1/3)
Therefore, event B and C are independent.
Hence, we have shown that event A and B are not independent, while event B and C are independent.
To show that Event A and B are independent, we need to prove that the probability of A and B occurring together is equal to the product of their individual probabilities.
Let's denote H as the event of getting a head and T as the event of getting a tail. Since the coin is unbiased, the probability of getting a head or a tail is both 0.5.
Event A: Head on the first toss and head on the second toss
The probability of getting a head on the first toss is 1/2, and on the second toss is also 1/2.
So, P(A) = P(H) * P(H) = 1/2 * 1/2 = 1/4.
Event B: Tail on the third toss
The probability of getting a tail on the third toss is 1/2.
So, P(B) = P(T) = 1/2.
Event A and B: Head on the first two tosses and tail on the third toss
The probability of getting a head on the first two tosses is 1/4 (as calculated above), and the probability of getting a tail on the third toss is 1/2.
So, P(A and B) = P(H) * P(H) * P(T) = 1/2 * 1/2 * 1/2 = 1/8.
To check if A and B are independent, we need to compare P(A and B) with P(A) * P(B):
P(A) * P(B) = (1/4) * (1/2) = 1/8.
Since P(A and B) = P(A) * P(B), we can conclude that Event A and B are independent.
Now, let's move on to proving that Event B and C are dependent.
Event C: Exactly two tails appear in the three tosses
To calculate the probability of Event C, we need to consider the different combinations of two tails in three tosses: TTH, THT, and HTT.
The probability of each combination is:
P(TTH) = P(T) * P(T) * P(H) = (1/2) * (1/2) * (1/2) = 1/8.
P(THT) = P(T) * P(H) * P(T) = (1/2) * (1/2) * (1/2) = 1/8.
P(HTT) = P(H) * P(T) * P(T) = (1/2) * (1/2) * (1/2) = 1/8.
So, P(C) = P(TTH) + P(THT) + P(HTT) = 1/8 + 1/8 + 1/8 = 3/8.
Event B and C: Tail on the third toss and exactly two tails in three tosses
To calculate the probability of Event B and C occurring together, we need to consider the combinations where a tail appears on the third toss and exactly two tails appear in three tosses.
The combinations are:
(TTH) - This satisfies Event B and C.
(THH) - This satisfies Event B but not C.
(THT) - This satisfies Event B and C.
So, P(B and C) = P(TTH) + P(THT) = 1/8 + 1/8 = 1/4.
Comparing P(B and C) with P(B) * P(C):
P(B) * P(C) = (1/2) * (3/8) = 3/16.
Since P(B and C) is not equal to P(B) * P(C), we can conclude that Event B and C are dependent.
Therefore, we have shown that Event A and B are independent, and Event B and C are dependent.