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Hint: If p and q be the probability of success and failure of an event respectively. Then the probability of that event getting ‘x’ success in an n-Bernoulli’s trial where (\[i \leqslant n\]), then \[P\left( {{A_x}} \right) = {}^n{C_x}{p^x}{q^{n - x}}\].
Complete step-by-step answer:
We do this problem using binomial law, which is, if Ax denotes the event of getting exactly x success in an n-Bernoulli’s trial where (\[i \leqslant n\]), then \[P\left( {{A_x}} \right) = {}^n{C_x}{p^x}{q^{n - x}}\], where p and q denotes the probability of success and failure of that event respectively and \[p + q = 1\].
Here, p = the probability that a person will hit the target in shooting practice = 0.3
And, q = the probability that a person will not hit a target in shooting practice\[ = 1 - p = 1 - 0.3 = 0.7\]
Here, the total number of shoots is 10 i.e. n=10.
\[\therefore \]The probability that out of 10 shoots none will hit a target\[ = {}^{10}{C_0}{p^0}{q^{10 - 0}} = {q^{10}} = {\left( {0.7} \right)^{10}}\].
Therefore, the probability of shooting target is \[1 - {\left( {0.7} \right)^{10}}\]
Hence, option (B) is correct.
Note: Another Method –
Let, A be the event that a person will hit a target in shooting practice.
Then the probability of A is given, which is \[P(A) = 0.3\].
Now, the probability that a person will not hit a target in shooting practice is \[P\left( {\bar A} \right) = 1 - P\left( A \right) = 1 - 0.3 = 0.7\]
[ We know \[P\left( A \right) + P\left( {\bar A} \right) = P\left( X \right) = 1\]\[i.e.,P\left( {\bar A} \right) = 1 - P\left( A \right)\] ]
Now, the probability that out of 10 shoots none will hit a target is \[P\left( {\bar A} \right),P\left( {\bar A} \right),.....upto\]10 times = \[{\left( {0.7} \right)^{10}}\]
Therefore, the probability of shooting the target =\[1 - {\left( {0.7} \right)^{10}}\]
Complete step-by-step answer:
We do this problem using binomial law, which is, if Ax denotes the event of getting exactly x success in an n-Bernoulli’s trial where (\[i \leqslant n\]), then \[P\left( {{A_x}} \right) = {}^n{C_x}{p^x}{q^{n - x}}\], where p and q denotes the probability of success and failure of that event respectively and \[p + q = 1\].
Here, p = the probability that a person will hit the target in shooting practice = 0.3
And, q = the probability that a person will not hit a target in shooting practice\[ = 1 - p = 1 - 0.3 = 0.7\]
Here, the total number of shoots is 10 i.e. n=10.
\[\therefore \]The probability that out of 10 shoots none will hit a target\[ = {}^{10}{C_0}{p^0}{q^{10 - 0}} = {q^{10}} = {\left( {0.7} \right)^{10}}\].
Therefore, the probability of shooting target is \[1 - {\left( {0.7} \right)^{10}}\]
Hence, option (B) is correct.
Note: Another Method –
Let, A be the event that a person will hit a target in shooting practice.
Then the probability of A is given, which is \[P(A) = 0.3\].
Now, the probability that a person will not hit a target in shooting practice is \[P\left( {\bar A} \right) = 1 - P\left( A \right) = 1 - 0.3 = 0.7\]
[ We know \[P\left( A \right) + P\left( {\bar A} \right) = P\left( X \right) = 1\]\[i.e.,P\left( {\bar A} \right) = 1 - P\left( A \right)\] ]
Now, the probability that out of 10 shoots none will hit a target is \[P\left( {\bar A} \right),P\left( {\bar A} \right),.....upto\]10 times = \[{\left( {0.7} \right)^{10}}\]
Therefore, the probability of shooting the target =\[1 - {\left( {0.7} \right)^{10}}\]
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