
A box contains tickets numbered 1 to N. n tickets are drawn at random with replacement. Find the probability that the largest number on the selected ticket is K, (K≤N)
A. ${\left( {\dfrac{K}{N}} \right)^n}$
B. ${\left( {\dfrac{K}{N}} \right)^{n - 1}}$
C. $\dfrac{{{K^n} - {K^{n - 1}}}}{{{N^n}}}$
D. $\dfrac{{{K^n} - {{\left( {K - 1} \right)}^n}}}{{{N^n}}}$
Answer
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Hint: Probability means possibility. It is a branch of mathematics that deals with the occurrence of events. It is used to predict how likely events are to happen. To find the probability of a single event to occur, first, we should know the total number of possible outcomes. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
Random experiment is an experiment where we know the set of all possible outcomes but find it impossible to predict one at any particular execution. Sample space is defined as the set of all possible outcomes of a random experiment. The probability of all the events in a sample space adds up to 1.
Event is a subset of the sample space i.e. a set of outcomes of the random experiment.
\[{\text{Probability of an event}} = \dfrac{{{\text{Number of occurence of event A in S}}}}{{{\text{Total number of cases in S}}}}{\text{ }}\]
\[ = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
Complete step by step solution: According to the question, the tickets are drawn with replacement; the same ticket may be repeated.
Since the largest ticket is k, the tickets are chosen from 1 to K
Therefore, total number of ways of doing it = ${K^n}$
However, K may or may not have been drawn
Number of ways of choosing numbers from 1 to (K-1) = ${\left( {K - 1} \right)^n}$
Thus, number of ways in which the maximum number drawn is K = $n(A) = {K^n} - {\left( {K - 1} \right)^n}$
And total number of ways = $n(S) = {N^n}$
Therefore, Probability = $\dfrac{{n(A)}}{{n(S)}}$
Probability = $\dfrac{{{K^n} - {{\left( {K - 1} \right)}^n}}}{{{N^n}}}$
∴Option (D) is correct
Note: There are some main rules associated with basic probability:
1. P(not A)=1-P(A)
2. P(A or B)=P(event A occurs or event B occurs or both occur)
3. P(A and B)=P(both event A and event B occurs)
4. The general Addition Rule: P(A or B) = P(A) + P(B) – P(A and B)
Random experiment is an experiment where we know the set of all possible outcomes but find it impossible to predict one at any particular execution. Sample space is defined as the set of all possible outcomes of a random experiment. The probability of all the events in a sample space adds up to 1.
Event is a subset of the sample space i.e. a set of outcomes of the random experiment.
\[{\text{Probability of an event}} = \dfrac{{{\text{Number of occurence of event A in S}}}}{{{\text{Total number of cases in S}}}}{\text{ }}\]
\[ = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
Complete step by step solution: According to the question, the tickets are drawn with replacement; the same ticket may be repeated.
Since the largest ticket is k, the tickets are chosen from 1 to K
Therefore, total number of ways of doing it = ${K^n}$
However, K may or may not have been drawn
Number of ways of choosing numbers from 1 to (K-1) = ${\left( {K - 1} \right)^n}$
Thus, number of ways in which the maximum number drawn is K = $n(A) = {K^n} - {\left( {K - 1} \right)^n}$
And total number of ways = $n(S) = {N^n}$
Therefore, Probability = $\dfrac{{n(A)}}{{n(S)}}$
Probability = $\dfrac{{{K^n} - {{\left( {K - 1} \right)}^n}}}{{{N^n}}}$
∴Option (D) is correct
Note: There are some main rules associated with basic probability:
1. P(not A)=1-P(A)
2. P(A or B)=P(event A occurs or event B occurs or both occur)
3. P(A and B)=P(both event A and event B occurs)
4. The general Addition Rule: P(A or B) = P(A) + P(B) – P(A and B)
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