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Question: There are 25 stamps numbered from 1 to 25 in a box. If a stamp is drawn at random from the box. The ...

There are 25 stamps numbered from 1 to 25 in a box. If a stamp is drawn at random from the box. The probability that the number on the stamp drawn is a prime number is
A. 1225\dfrac{{12}}{{25}}
B. 1325\dfrac{{13}}{{25}}
C. 925\dfrac{9}{{25}}
D. 625\dfrac{6}{{25}}

Explanation

Solution

At first we’ll find the number of ways of selection any 1 out of 25 stamps, then we’ll find the numbers of prime numbers from 1 to 25 and the number of ways of selecting one prime number out of all prime number. Then using the formula for probability, i.e., probability of an event =favourable outcomestotal possible outcomes = \dfrac{{favourable{\text{ outcomes}}}}{{total{\text{ possible outcomes}}}}, we’ll find the required probability.

Complete step by step Answer:

Given data: Total number of stamps in the box=25
We know that the number of ways of selecting any ‘r’ elements out of a total ‘n’ number of elements irrespective of the order of ‘r’ elements is given by nCr{}^n{C_r},
Where, nCr=n!r!(nr)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}and n!=n(n1)(n2)(n3)(n4)..........(3)(2)(1)n! = n(n - 1)(n - 2)(n - 3)(n - 4)..........(3)(2)(1)
Therefore the number of ways that any 1 stamp is chosen out of 25=25C1 = {}^{25}{C_1}
Prime numbers from 1 to 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23
Therefore, the number of prime numbers between 1 and 25=9
The number of ways that any prime numbered stamp is chosen out of 9 primes=9C1 = {}^9{C_1}
As, probablilty of an event =favourable outcomestotal possible outcomes = \dfrac{{favourable{\text{ outcomes}}}}{{total{\text{ possible outcomes}}}}
Therefore, the probability that the number on the stamp drawn is a prime number=9C125C1 = \dfrac{{{}^9{C_1}}}{{{}^{25}{C_1}}}
Using nCr=n!r!(nr)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, we get,
=9!1!(91)!25!1!(251)!= \dfrac{{\dfrac{{9!}}{{1!\left( {9 - 1} \right)!}}}}{{\dfrac{{25!}}{{1!\left( {25 - 1} \right)!}}}}
Now using n!=n(n1)!n! = n\left( {n - 1} \right)! , we get,
=9×8!1!(8)!25×24!1!(24)!= \dfrac{{\dfrac{{9 \times 8!}}{{1!\left( 8 \right)!}}}}{{\dfrac{{25 \times 24!}}{{1!\left( {24} \right)!}}}}
On cancelling common terms we get,
=925= \dfrac{9}{{25}}
Hence, the probability that the number on the stamp drawn is a prime number is 925\dfrac{9}{{25}}

Note: Here we have taken prime numbers like 2, 3, 5 and, 7, but some students may count 1 as well as it satisfies both the condition according to the definition of a prime number which is, A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. But there is also a definition of prime number that has only two factors which are not satisfied by 1, hence it is not included under the category of prime numbers.