When a die is rolled and a coin(with heads and tails) is tossed,find the probability of obtaining,
(a). Tails and an even number.
solution.
A={HH,TT,HT,HT}
B={HH,TH}
(b). A number greater than 3
SOLUTION.
let event C=(AuB)
P(C)=P(AuB)=P(A)+P(B)-P(AnB)
=3/4
To find the probability of obtaining tails and an even number when a die is rolled and a coin is tossed, we need to find the intersection of the event "tails" and the event "even number".
The event A represents the outcomes of the coin toss and the die roll: A = {HH, TT, HT, HT}. The event B represents the outcomes of obtaining tails: B = {HH, TH}.
(a) To find the probability of obtaining tails and an even number, we need to find the intersection of event A and event B. In other words, we need to find the common outcomes of A and B: A∩B = {HH}.
The probability of obtaining tails and an even number is given by: P(A∩B) / P(S), where P(S) represents the sample space, or the total number of possible outcomes.
In this case, there is only one outcome in the intersection of A and B, and the sample space consists of 4 possible outcomes.
Therefore, the probability of obtaining tails and an even number is: P(A∩B) / P(S) = 1 / 4.
(b) To find the probability of obtaining a number greater than 3, we need to find the union of event A and event B. In other words, we need to find the combined outcomes of A and B: A∪B = {HH, TT, HT, TH}.
The probability of obtaining a number greater than 3 is given by: P(A∪B) / P(S).
In this case, there are 4 outcomes in the union of A and B, and the sample space consists of 4 possible outcomes.
Therefore, the probability of obtaining a number greater than 3 is: P(A∪B) / P(S) = 4 / 4 = 1.
To find the probability of obtaining tails and an even number, we need to find the intersection of the events "tails" and "even number".
Given that the possible outcomes when rolling a die are {1, 2, 3, 4, 5, 6}, and the possible outcomes when tossing a coin are {heads, tails}, we can represent the sample space as a cartesian product of the two sets: {(1, heads), (1, tails), (2, heads), (2, tails), (3, heads), (3, tails), (4, heads), (4, tails), (5, heads), (5, tails), (6, heads), (6, tails)}.
The event "tails" can be represented as {(1, tails), (2, tails), (3, tails), (4, tails), (5, tails), (6, tails)}, and the event "even number" can be represented as {(2, heads), (2, tails), (4, heads), (4, tails), (6, heads), (6, tails)}.
The intersection of these two events is {(2, tails), (4, heads), (4, tails), (6, heads), (6, tails)}, which is represented as "HT, HH, HT, HH, HT" in simplified form.
Therefore, the probability of obtaining tails and an even number is given by the ratio of the number of favorable outcomes to the total number of outcomes.
P(tails and even number) = 5 (number of favorable outcomes) / 12 (total number of outcomes) = 5/12.
(b) To find the probability of obtaining a number greater than 3, we can list the possible outcomes: {4, 5, 6}.
The event "number greater than 3" can be represented as {(4, heads), (4, tails), (5, heads), (5, tails), (6, heads), (6, tails)}.
The number of outcomes in this event is 6, and since there are 12 total outcomes, the probability of obtaining a number greater than 3 is:
P(number greater than 3) = 6 (number of favorable outcomes) / 12 (total number of outcomes) = 6/12 = 1/2.