Question

A boy is throwing stones at a target. The probability of hitting the target at any trial is $\dfrac{1}{2}$ . The probability of hitting the target 5th time at the 10th row is
A) $\dfrac{5}{{{2^{10}}}}$
B) $\dfrac{{63}}{{{2^9}}}$
C) $\dfrac{{{}^{10}{C_5}}}{{{2^{10}}}}$
D) $\dfrac{{{}^{10}{C_4}}}{{{2^{10}}}}$

Answer 1

Hint : Hitting the target 5th time at the 10th row means hitting the target 4 times in the previous 9 rows. So find the probability of hitting the target 4 times in 9 rows and multiply it with the probability of hitting the target at 10th row. Probability of hitting the target r times in n rows is ${}^n{C_r}{x^r}{y^{n - r}}$ , where x is the probability of success and y is the probability of failure. Use this formula to find the probability.

Complete step-by-step answer :
We are given that a boy is throwing stones at a target.
Given, the probability of hitting the target x is $\dfrac{1}{2}$
Therefore, the probability of not hitting the target y will be
$1 - x \\\ = 1 - \dfrac{1}{2} \\\ = \dfrac{1}{2} \\\$
Probability of hitting the target 5th time at 10th row (P) = Probability of hitting the target 4 times in 9 rows × Probability of just hitting the target at 10th row
Probability of just hitting the target at 10th row will be $\dfrac{1}{2}$
$P = \left( {{}^n{C_r}{x^r}{y^{n - r}}} \right)\left( {\dfrac{1}{2}} \right) \\\ n = 9,r = 4,x = \dfrac{1}{2},y = \dfrac{1}{2} \\\ P = \left( {{}^9{C_4}{{\left( {\dfrac{1}{2}} \right)}^4}{{\left( {\dfrac{1}{2}} \right)}^{9 - 4}}} \right)\left( {\dfrac{1}{2}} \right) \\\ P = {}^9{C_4}{\left( {\dfrac{1}{2}} \right)^4}{\left( {\dfrac{1}{2}} \right)^5}\left( {\dfrac{1}{2}} \right) \\\ P = {}^9{C_4}{\left( {\dfrac{1}{2}} \right)^{10}} \\\ {}^9{C_4} = \dfrac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}} = \dfrac{{3024}}{{24}} = 126 \\\ P = 126 \times \dfrac{1}{{{2^{10}}}} \\\ P = \dfrac{{126}}{{{2^{10}}}} = \dfrac{{63}}{{{2^9}}} = \dfrac{{63}}{{512}} \\\$
Therefore, the probability of hitting the target 5th time at the 10th row is $\dfrac{{63}}{{{2^9}}}$
From among the options given in the question Option B is correct, which is $\dfrac{{63}}{{{2^9}}}$
So, the correct answer is “Option B”.

Note : If probability of an Event E is x then the probability of E’ (compliment) will be $1 - x$. Whenever we are not sure about the outcome of an event, we can talk about the probabilities of certain outcomes and how likely they are. The sum of probabilities of all possible outcomes is 1. The probabilities of an event and its complement must always total to 1.