A coin is tossed three times. Find the probability of each event.
a) at least two tosses come up tails.
b) at least one toss comes up heads.
PLEASE SHOW ALL WORK!
a) 2/3
b) 1/3
There are only 8 possible outcomes:
HHH, HTT, THT, TTH, HHT, HTH, THH, TTT
Of those, there are 4 with at least two tails
a) prob = 4/8 = 1/2
b) There are 7 with at least one H, (only TTT has no heads)
so prob = 7/8
To find the probability of each event, we need to determine the total number of possible outcomes and the number of favorable outcomes for each event.
Let's start with event a: "at least two tosses come up tails."
1. Total number of possible outcomes:
When a coin is tossed, there are two possible outcomes: heads (H) or tails (T). Since the coin is tossed three times, the total number of possible outcomes is 2^3 = 8.
2. Number of favorable outcomes:
In this case, we need to find the number of outcomes where at least two tosses come up tails. This includes the following scenarios:
- 3 tails: TTT
- 2 tails, 1 head: TTH, THT, HTT
Therefore, there are 4 favorable outcomes.
3. Probability of event a:
The probability of an event is given by the ratio of favorable outcomes to the total number of outcomes. So, the probability of event a is 4/8 = 1/2.
Moving on to event b: "at least one toss comes up heads."
1. Total number of possible outcomes:
Since we already calculated the total number of possible outcomes in the previous question, it remains 8.
2. Number of favorable outcomes:
In this case, we need to find the number of outcomes where at least one toss comes up heads. This includes the following scenarios:
- 3 heads: HHH
- 2 heads, 1 tail: HHT, HTH, THH
- 1 head, 2 tails: HTT, THT, TTH
Therefore, there are 7 favorable outcomes.
3. Probability of event b:
Similar to event a, we can calculate the probability of event b as 7/8.
So, the probability of event a is 1/2 or 50%, and the probability of event b is 7/8 or 87.5%.