Question

An unbaised coin is tossed ' n ' times. Let the random variable X denote the number of times the head occurs. If P ( X = 1 ), P ( X = 2 ), P ( X = 3 ) are in A.P. Find the value of n.

Answer 1

Answer:

Step-by-step explanation:The probability of exactly k success with success probability p in n trials is given by:(nk)∗pk∗(1−p)n−k

.

For an unbiased coin, probability of getting head is: p=1−p=0.5

Hence, probability of exactly k

heads in n trials is : (nk)∗(0.5)k∗(0.5)n−k=(nk)∗(0.5)n

So, (n1)∗(0.5)n,(n2)∗(0.5)nand(n3)∗(0.5)n

are in AP.

An AP divided by a constant number would also be in AP. So, we can easily get rid of (0.5)n

. Thus, (n1),(n2)and(n3)

are in AP.

So,

(n1)+(n3)=2∗(n2)

(n3−3n2+2n)/6+n=2∗(n2−n)/2

=>(n3−3n2+8n)/6=(n2−n)

=>n=0

or (n2−3n+8)=6(n−1)

=>n=0

or n2−9n+14=0

=>n=0

or n=2 or n=7

Obviously, since we are getting 3 heads, first two answers are invalid. So, n=7

.