an unbiased coin is tossed six times. find the probability of the given event - the coin lands heads more than once.
must exclude no heads and 1 head
prob(no heads) = (1/2)^6 = 1/64
prob(1 head) = C(6,1) (1/2)^1 (1/2)^5 = 6/64
prob (more than 1 head) = 1 - 1/64 - 6/64
= 57/64
thank you so much, this helped alot
To find the probability of the event "the coin lands heads more than once" when a coin is tossed six times, we can use the concept of binomial probability.
The probability of getting a specific outcome (in this case, heads) in a single coin toss is 0.5 since the coin is unbiased.
Let's calculate the probability for each possible outcome:
Getting one head:
The probability of getting one head and five tails is given by the binomial coefficient formula:
P(1 head) = (6C1) * (0.5)^1 * (0.5)^5 = 6 * 0.5^6 = 6/64 = 3/32
Getting two heads:
P(2 heads) = (6C2) * (0.5)^2 * (0.5)^4 = 15 * 0.5^6 = 15/64
Getting three heads:
P(3 heads) = (6C3) * (0.5)^3 * (0.5)^3 = 20 * 0.5^6 = 20/64 = 5/16
To find the probability of getting more than one head, we sum up the probabilities of getting two or three heads:
P(more than one head) = P(2 heads) + P(3 heads)
= 15/64 + 5/16
= 15/64 + 20/64
= 35/64
Therefore, the probability of getting more than one head when tossing an unbiased coin six times is 35/64.
To find the probability of an event, we need to determine the total number of possible outcomes and the number of outcomes that satisfy the given condition.
In this problem, we have an unbiased coin that is tossed six times, and we want to find the probability of the event where the coin lands heads more than once.
To compute the total number of possible outcomes, we need to consider that each coin toss has two possible outcomes: heads or tails. Since we have six tosses, the total number of possible outcomes is 2^6 = 64.
Now let's determine the number of outcomes that satisfy the condition of the event, where the coin lands heads more than once. We can calculate this by considering the possible number of heads out of the six tosses.
If we have one head, there are six possible positions for it:
- H _ _ _ _ _
- _ H _ _ _ _
- _ _ H _ _ _
- _ _ _ H _ _
- _ _ _ _ H _
- _ _ _ _ _ H
Similarly, if we have two heads, there are fifteen possible positions for them:
- H H _ _ _ _
- H _ H _ _ _
- H _ _ H _ _
- H _ _ _ H _
- H _ _ _ _ H
- _ H H _ _ _
- _ H _ H _ _
- _ H _ _ H _
- _ H _ _ _ H
- _ _ H H _ _
- _ _ H _ H _
- _ _ H _ _ H
- _ _ _ H H _
- _ _ _ H _ H
- _ _ _ _ H H
Thus, the number of outcomes that satisfy the condition of the event is 6 + 15 = 21.
Now, we have all the information we need to determine the probability. The probability of an event is given by the ratio of the number of outcomes that satisfy the event to the total number of possible outcomes.
Therefore, the probability of the coin landing heads more than once is 21/64, which can be simplified, if needed.