A key bunch contains m keys exactly one of which opens the door. A man tries to open the door by choosing a key at random discarding the wrong key after each trial. The probability that the door opens at the r’th trial (1≤r≤m), is
a. $\dfrac{{\text{1}}}{{\text{m}}}$
b. $\dfrac{{\text{1}}}{{\text{r}}}$
c. $\dfrac{{\text{r}}}{{\text{m}}}$
d. $\dfrac{{\left( {{\text{r - 1}}} \right)}}{{\text{m}}}$
Answer
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Hint: To solve this question, we need to use the basic theory related to the probability. As we know, to calculate the probability of an event, we take the number of successes, divided by the number of possible outcomes.
Complete step-by-step answer:
The probability of event A happening is:
P(A) = $\dfrac{{{\text{n}}\left( {\text{A}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}}$
Where,
P(A) $ \to $ probability of event A happening
n(A) $ \to $number of favorable outcomes
n(S) $ \to $ total number of outcomes
Now, as given in question,
A key bunch contains m keys exactly one of which opens the door and also a man tries to open the door by choosing a key at random discarding the wrong key after each trial.
Therefore, Probability that it will open in the first try = $\dfrac{{\text{1}}}{{\text{m}}}$
Probability that it will open in the second try = $\dfrac{{{\text{m - 1}}}}{{\text{m}}}{\times }\dfrac{{\text{1}}}{{{\text{m - 1}}}}$
= $\dfrac{{\text{1}}}{{\text{m}}}$
Probability that it will open in the third try = $\dfrac{{{\text{m - 1}}}}{{\text{m}}}{\times }\dfrac{{{\text{m - 2}}}}{{{\text{m - 1}}}}{ \times }\dfrac{{\text{1}}}{{{\text{m - 2}}}}$
= $\dfrac{{\text{1}}}{{\text{m}}}$
Thus, the probability of opening is always $\dfrac{{\text{1}}}{{\text{m}}}$irrespective of the number of trials.
Therefore, the probability that the door opens at the r’th trial (1≤r≤m), is $\dfrac{{\text{1}}}{{\text{m}}}$.
Thus, option (A) is the correct answer.
Note- Probability is a number between 0 and 1. We multiply our final answer by 100 to get a “percent”. We borrowed that concept from the French - “per cent” - which means per 100. Something with 0 or 0% probability will never happen. Something with 1 or 100% percent probability will always happen. Something with 0.5 or 50% probability will happen half the time.
If your final answer is smaller than 0 or bigger than 1, then you made a mistake. Everything happens between 0 and 1.
Complete step-by-step answer:
The probability of event A happening is:
P(A) = $\dfrac{{{\text{n}}\left( {\text{A}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}}$
Where,
P(A) $ \to $ probability of event A happening
n(A) $ \to $number of favorable outcomes
n(S) $ \to $ total number of outcomes
Now, as given in question,
A key bunch contains m keys exactly one of which opens the door and also a man tries to open the door by choosing a key at random discarding the wrong key after each trial.
Therefore, Probability that it will open in the first try = $\dfrac{{\text{1}}}{{\text{m}}}$
Probability that it will open in the second try = $\dfrac{{{\text{m - 1}}}}{{\text{m}}}{\times }\dfrac{{\text{1}}}{{{\text{m - 1}}}}$
= $\dfrac{{\text{1}}}{{\text{m}}}$
Probability that it will open in the third try = $\dfrac{{{\text{m - 1}}}}{{\text{m}}}{\times }\dfrac{{{\text{m - 2}}}}{{{\text{m - 1}}}}{ \times }\dfrac{{\text{1}}}{{{\text{m - 2}}}}$
= $\dfrac{{\text{1}}}{{\text{m}}}$
Thus, the probability of opening is always $\dfrac{{\text{1}}}{{\text{m}}}$irrespective of the number of trials.
Therefore, the probability that the door opens at the r’th trial (1≤r≤m), is $\dfrac{{\text{1}}}{{\text{m}}}$.
Thus, option (A) is the correct answer.
Note- Probability is a number between 0 and 1. We multiply our final answer by 100 to get a “percent”. We borrowed that concept from the French - “per cent” - which means per 100. Something with 0 or 0% probability will never happen. Something with 1 or 100% percent probability will always happen. Something with 0.5 or 50% probability will happen half the time.
If your final answer is smaller than 0 or bigger than 1, then you made a mistake. Everything happens between 0 and 1.
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