Probability of hitting the target Nth time at Mth throw | By The Digital Insider

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Given integers N, M and p, the task is to find the probability of hitting a target for the N^th time at the M^th throw where p is the probability of hitting the target.

Examples:

Input: N = 1, M = 2, p = 0.3
Output: 0.21
Explanation: The target can be hit for the first time in the second throw only when the first throw is a miss.
So, the probability is multiplication of probability of hitting the target in second throw and probability of missing the target in the first throw = 0.3 * 0.7 = 0.21
because probability of missing  = 1 – probability of hitting.

Input: N = 8, M = 17, p = 0.4,
Output: 0.07555569565040642

Approach: The idea to solve the problem is based on the binomial distribution of probability

For getting N^th hit in the M^th throw there must be N-1 hits in the M-1 thrwos earlier and Nth throw must be a hit.
Say p is the probability of hitting and q is the probability of missing. q = 1 – p.
So the probability of getting N-1 hits in M-1 throws is: 
X = ^M-1C_N-1 (p^N-1q^M-1-(N-1)) = ^M-1C_N-1 (p^N-1q^M-N)
Therefore, the probability of hitting for N^th time is p*X = ^M-1C_N-1 (p^Nq^M-N)

Follow the below steps to implement the above idea:

• Firstly, get the probability of not hitting a target.


• Get the value of X as shown above.


• Multiply the value of ‘p’ with it to get the actual answer.

Below is the implementation of the above approach.

Python3

def probab(p, m, n):     q = 1-p     res = fact(m-1)/fact(n-1)/fact(m-n)*p**(n-1)*q**(m-n)*p     return res    def fact(f):     fact = 1     for i in range(1, f + 1):         fact = i * fact     return fact       if __name__ == '__main__':     p = 0.3     M = 2     N = 1     print(probab(p, M, N))

Output

0.21

Time Complexity: O(M)
Auxiliary Space: O(1)

#Code, #Mathematical, #Probability

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Published on The Digital Insider at https://bit.ly/3GuUUpe.