
In a lottery, a person choses six different natural numbers at random from 1 to 20 and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [Hint: order of the numbers is not important]
Answer
624.3k+ views
Hint: To solve the question, we have to analyse the number of ways the combinations can be chosen. To calculate the number of ways of choosing six numbers from the given set of 20 numbers, apply the formulae of combinations. We have to analyse that only one chance of winning exists since only one set of matching sets of six numbers exists. To calculate the probability, use the concept of ratio of chance of winning to total number of chances.
Complete step by step answer:
The given number of numbers present in the lottery is equal to 6.
The set of numbers from which the given 6 numbers can be chosen is equal to 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.
The total numbers in the given set is equal to 20.
The number of ways of choosing 6 numbers from the given set of 20 numbers
= \[{}^{20}{{C}_{6}}\]
We know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] and \[n!=n\left( n-1 \right)\left( n-2 \right).....2\times 1\]
By applying the above formula, we get
\[\begin{align}
& {}^{20}{{C}_{6}}=\dfrac{20!}{6!\left( 20-6 \right)!} \\
& =\dfrac{20!}{6!\left( 14 \right)!} \\
& =\dfrac{20\times 19\times 18\times 17\times 16\times 15\times 14!}{6!\left( 14 \right)!} \\
& =\dfrac{20\times 19\times 18\times 17\times 16\times 15}{6!} \\
& =\dfrac{20\times 19\times 18\times 17\times 16\times 15}{6\times 5\times 4\times 3\times 2\times 1} \\
& =19\times 17\times 8\times 15 \\
& =323\times 120 \\
& =38760 \\
\end{align}\]
The number of ways six numbers match with the six numbers already fixed by the lottery committee is equal 1 way.
Since only one set of matching sets of six numbers exits.
The probability of winning the prize in the game = Ratio of the number of ways six numbers match with the six numbers already fixed by the lottery committee, to the number of ways of choosing 6 numbers from the given set of 20 numbers
\[=\dfrac{1}{38760}\]
\[\approx 0.000026\]
Thus, the probability of winning the prize in the game is equal to 0.000026
Note: The possibility of mistake can be, not analysing that only one chance of winning exists since only one set of matching sets of six numbers exists. The other possibility of mistake can be, not applying the formulae of combinations to ease the procedure of solving.
Complete step by step answer:
The given number of numbers present in the lottery is equal to 6.
The set of numbers from which the given 6 numbers can be chosen is equal to 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.
The total numbers in the given set is equal to 20.
The number of ways of choosing 6 numbers from the given set of 20 numbers
= \[{}^{20}{{C}_{6}}\]
We know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] and \[n!=n\left( n-1 \right)\left( n-2 \right).....2\times 1\]
By applying the above formula, we get
\[\begin{align}
& {}^{20}{{C}_{6}}=\dfrac{20!}{6!\left( 20-6 \right)!} \\
& =\dfrac{20!}{6!\left( 14 \right)!} \\
& =\dfrac{20\times 19\times 18\times 17\times 16\times 15\times 14!}{6!\left( 14 \right)!} \\
& =\dfrac{20\times 19\times 18\times 17\times 16\times 15}{6!} \\
& =\dfrac{20\times 19\times 18\times 17\times 16\times 15}{6\times 5\times 4\times 3\times 2\times 1} \\
& =19\times 17\times 8\times 15 \\
& =323\times 120 \\
& =38760 \\
\end{align}\]
The number of ways six numbers match with the six numbers already fixed by the lottery committee is equal 1 way.
Since only one set of matching sets of six numbers exits.
The probability of winning the prize in the game = Ratio of the number of ways six numbers match with the six numbers already fixed by the lottery committee, to the number of ways of choosing 6 numbers from the given set of 20 numbers
\[=\dfrac{1}{38760}\]
\[\approx 0.000026\]
Thus, the probability of winning the prize in the game is equal to 0.000026
Note: The possibility of mistake can be, not analysing that only one chance of winning exists since only one set of matching sets of six numbers exists. The other possibility of mistake can be, not applying the formulae of combinations to ease the procedure of solving.
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