Question
Question: A coin is tossed \[n\] times. The probability of getting at least one head is greater than that of g...
A coin is tossed n times. The probability of getting at least one head is greater than that of getting at least two tails by 325. Then n is:
(A) 5
(B) 10
(C) 15
(D) None of these
Solution
According to the question, we need to know which formula should be used. Here, we have to use the binomial distribution formula. After applying the formula for both the cases, we need to solve it by the help of the given relationship between the two cases in the question
Complete step-by-step solution:
We know that we have to solve this question by the binomial distribution formula. The formula is:
⇒nCmpmqn
Here, in this formula, nis the number of trials, mis the number of successes, pis probability of success, qis the probability of failure.
We know that when we toss a coin, then the probability of getting one head is 21, and the probability of getting a tail is 21.
Now, the question is the probability of getting at least one head. If we take the probability of getting at least one head as P(A), then we can represent P(A)as:
P(A)=1−P(noheads)
Now, when we further solve this. Here, we will put the binomial distribution formula. When we put the formula in the above given equation, we get that m=0because it is the probability of getting no heads. Then, pis the probability of success, this means the probability of getting a head, and the probability of getting a head is 21. So, the value of pis 21. Now, qis the probability of failure. If the probability of success is 21, the probability of failure is 21. So, the value of qis 21. Now, when we put the values in the equation, we get:
P(A)=1−nC0(21)0(21)n
We know that the value of nC0is 1. When we solve this, we get:
⇒P(A)=1−(21)n
Now, the next case is the probability of getting two tails. We can represent it as P(B). Now, when we find the probability of getting at least two tails, we get:
P(B)=1−P(notail)−P(1tail)
When we solve this according to the formula, we get that m=0because it is the probability of getting no tail. Then, pis the probability of success, this means the probability of getting a tail, and the probability of getting a tail is 21. So, the value of pis 21. Now, qis the probability of failure. If the probability of success is 21, the probability of failure is 21. So, the value of qis 21. Now, for the probability of getting one tail, we get m=1and n−1. Now, when we put the values in the equation, we get:
P(B)=1−nC0(21)0(21)n−nC1(21)1(21)n−1
We know that the value of nC0is 1 and the value of nC1is n. When we solve this, we get:
⇒P(B)=1−(21)n−n(21)n
Now, according to the question, we get that the relationship between both P(A)and P(B)is:
⇒P(A)−P(B)=325
Now, we will put the values of P(A)and P(B), and we get:
⇒1−(21)n−[1−(21)n−n(21)n]=325
⇒1−(21)n−1+(21)n+n(21)n=325
⇒n(21)n=325
⇒2nn=325
When we compare both the sides, we get n=5.
Thus the correct answer is option ‘A’.
Note: In Mathematics, binomial distribution is a probability distribution of the number of successes and the number of failures. This distribution consists of parameters like nandp. This distribution is very common and is used in statistics.