What is the probability that when a fair coin is tossed 2 times then at least one of them are heads?
3/4
When solving probability questions involving coin tosses, it's essential to list all possible outcomes and then identify the specific outcomes that meet the given condition. This question asks for the probability of getting at least one head when a fair coin is tossed 2 times.
Coin Toss Outcomes Explained
A fair coin means that each side (head or tail) has an equal chance of appearing in a toss. When a fair coin is tossed 2 times, we need to consider all the possible combinations for the two tosses. Let 'H' represent a Head and 'T' represent a Tail.
- The first toss can be a Head (H) or a Tail (T).
- The second toss can also be a Head (H) or a Tail (T).
Combining these possibilities, the total possible outcomes when tossing a fair coin 2 times are:
| Toss 1 | Toss 2 | Combined Outcome |
|---|---|---|
| Head (H) | Head (H) | HH |
| Head (H) | Tail (T) | HT |
| Tail (T) | Head (H) | TH |
| Tail (T) | Tail (T) | TT |
From this table, we can see that there are 4 total possible outcomes when a fair coin is tossed 2 times.
Favorable Outcomes for At Least One Head
The question asks for the probability of getting at least one head. This phrase means that we are looking for outcomes where there is one head OR two heads. It excludes the case where there are no heads (i.e., both are tails).
Let's look at our list of total outcomes and identify the ones that have at least one head:
- HH: Contains two heads (satisfies "at least one head").
- HT: Contains one head (satisfies "at least one head").
- TH: Contains one head (satisfies "at least one head").
- TT: Contains no heads (does NOT satisfy "at least one head").
Therefore, the favorable outcomes for getting at least one head are HH, HT, and TH.
The number of favorable outcomes is 3.
Calculating Probability of At Least One Head
The formula for probability is:
$$\text{Probability (Event)} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$$
In this problem:
- Number of Favorable Outcomes (getting at least one head) = 3
- Total Number of Possible Outcomes (when tossing a fair coin 2 times) = 4
Now, substitute these values into the probability formula:
$$\text{P(at least one head)} = \frac{3}{4}$$
So, the probability that when a fair coin is tossed 2 times, then at least one of them are heads, is 3/4.
Important Questions from Conditional Probability
Two events A and B are such that P(not B) = 0.8, P(A ∪ B) = 0.5 and P(A|B) = 0.4. Then P(A) is equal to
For two events, A and B, it is given that \({\rm{P}}\left( {\rm{A}} \right) = \frac{3}{5},{\rm{\;P}}\left( {\rm{B}} \right) = \frac{3}{{10}}\) and \({\rm{P}}\left( {{\rm{A|B}}} \right) = \frac{2}{3}\) . If A̅ and B̅ are the complementary events of A and B, then what is P(A̅ | B̅) equal to?
For two mutually exclusive events A and B, P(A) = 0.2 and P (A̅ ∩ B) = 0.3. What is P (A|(A ∪ B)) equal to?
If an event B has occurred and has P(B) = 1, the conditional probability P(A|B) is equal to:
If P(A) = 0.7, P(B) = 0.5 and P(B/A) = 0.3, find (i) P(A/B) (ii) P(A ∪ B)?
