Question

A fair coin is tossed 15 times. The probability that the tail will appear atleast thrice is​

Answer 1

Answer:

In the last five throws there can be 0,1,2,3,4 or 5 heads and the same should be the case in the first ten throws.

Hence the favourable number of cases m=5C010C0+5C110C1+5C210C2+5C310C3+5C410C4+5C510C5

⇒m=1+50+450+1200+1050+252=3003.

Total number of ways n=215=32768.

Hence the required probability=nm=327683003

Answer 2

Concept

Since we know that when the two or more probability are independent of each other then they are multiplied and when it is dependent on each other then all the probability get added. Since the coin is tossed 15 times and it is said that at least three times the tail appears, then we will add the probability of getting at least three time tails with the probability of getting four time tails till the probability of getting tails by 15 times.

Given

The coins are tossed 15 times.

Find

We have to calculate the probability of getting a tail thrice.

Solution

Since we know that when a coin is tossed then the probability of getting either tail or head is 1/2. And when the coins are tossed then the probability for each time is independent of each other. Therefore probability of getting tail three time,

P1=(1/2)^3

Similarly the probability of getting tail four times, P2=(1/2)^4

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The probability of getting 15 times tail P13=(1/2)^15

Since all these different probabilities are dependent then the total probability can be written as

P=(1/2)^3 + (1/2)^4 + ……….+(1/2)^15

This is the geometric series and the formula to calculate the sum is given as,

Sum = a(1-r^n)/(1-r)

Here, n=13, r=1/2 and a=(1/2)^3

Therefore,

P= (1/2)^3[ 1- (1/2)^13]/[1- 1/2]

P= (1/2)^2[ 1- (1/2)^13]

Hence the probability of getting tail at least thrice in 15 throws is (1/2)^2[ 1- (1/2)^13].

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