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# Likelihood of hitting the goal Nth time at Mth throw

Given integers N, M and p, the task is to find the probability of hitting a target for the Nth time at the Mth 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 Nth hit in the Mth 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-1CN-1 (pN-1qM-1-(N-1)) = M-1CN-1 (pN-1qM-N)
Therefore, the probability of hitting for Nth time is p*X = M-1CN-1 (pNqM-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))`

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

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