How many tosses of a coin are needed so that the probability of getting at least one head is 0.875?

Question

Answer

Probability of getting at least one head in coin toss

In order to find out how many coin tosses are needed to get at least one head with a probability of 0.875, we need to understand the concept of probability and its calculation. Probability is the measure of the likelihood of an event occurring. It is expressed as a number between 0 and 1, where 0 represents an impossible event and 1 represents a certain event.

Calculation of probability

The probability of getting at least one head in a coin toss can be calculated using the formula:

P(at least one head) = 1 - P(no head)

Where P(no head) is the probability of getting no head in a coin toss.

The probability of getting no head in a single coin toss is:

P(no head) = 1/2

Therefore, the probability of getting at least one head in a single coin toss is:

P(at least one head) = 1 - P(no head) = 1 - 1/2 = 1/2

Calculation of number of coin tosses

In order to find out how many coin tosses are needed to get at least one head with a probability of 0.875, we can use the following formula:

P(at least one head) = 1 - (1/2)^n

Where n is the number of coin tosses.

Substituting the given probability value, we get:

0.875 = 1 - (1/2)^n

Solving for n, we get:

n = log(0.125) / log(0.5) = 3

Therefore, the number of coin tosses needed to get at least one head with a probability of 0.875 is 3.

Conclusion

In conclusion, we have seen that the probability of getting at least one head in a single coin toss is 1/2. Using this probability, we have calculated the number of coin tosses needed to get at least one head with a probability of 0.875 to be 3.

Answer

In the tossing of three head
Sample space={ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }
n(s) =8
Event A= Getting atleast one head
A={HHH, HHT, HTH, HTT, THH, THT, TTH}
n(a) = 7
n(A)
p(A)= ——— = 7/8=0.875
n(s)

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