Question

In a lottery of 50 tickets numbered 1 to 50, two tickets are drawn simultaneously. The probability that both the tickets drawn have prime numbers.

& \text{A}\text{. }\dfrac{1}{35} \\\ & \text{B}\text{. }\dfrac{16}{35} \\\ & \text{C}\text{. }\dfrac{7}{35} \\\ & \text{D}\text{. }\dfrac{3}{35} \\\ \end{aligned}$$

Answer 1

From the question, it was given that in a lottery of 50 tickets numbered 1 to 50, two tickets are drawn simultaneously. We should find the probability that both the tickets are drawn have prime numbers. So, it is clear we should pick 2 prime numbers from 50 numbers. We know that the ratio of the number of favorable outcomes to the total number of outcomes is called a probability. Now we have to find the total number of favorable outcomes and we should find the total number of outcomes. Let us assume the total number of favorable outcomes are equal to F and the total number of outcomes is equal to N. Let us assume the probability is equal to P. It is clear that the ratio of F and N gives us the value of P.

Complete step-by-step solution:
From the question, it was given that in a lottery of 50 tickets numbered 1 to 50, two tickets are drawn simultaneously. We should find the probability that both the tickets are drawn have prime numbers.
So, it is clear we should pick 2 prime numbers from 50 numbers.
We know that the ratio of the number of favorable outcomes to the total number of outcomes is called probability.
Here, the number of favorable outcomes is equal to the number of ways to pick 2 prime numbers from all prime numbers from 1 to 50.
Among 1 to 50, the prime numbers are 2,3,5,7,11,13,17,19,23,29,31,37,41,43 and 47.
So, it is clear that there are 15 prime numbers among 1 to 50.
Now we should find the number of ways to pick 2 prime numbers among 15 prime numbers.
We know that the number of ways to pick k objects among n objects is equal to $^{n}{{C}_{r}}$.
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
So, let us assume the number of ways to pick 2 prime numbers among 15 prime numbers is equal to F.
So, let us substitute the value of n is equal to 15 and the value of r is equal to 2 in $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.

& \Rightarrow F{{=}^{15}}{{C}_{2}} \\\ & \Rightarrow F=\dfrac{15!}{2!\left( 15-2 \right)!} \\\ & \Rightarrow F=\dfrac{15!}{2!13!} \\\ & \Rightarrow F=\dfrac{\left( 15 \right)\left( 14 \right)\left( 13! \right)}{2!13!} \\\ & \Rightarrow F=\dfrac{\left( 15 \right)\left( 14 \right)}{2} \\\ & \Rightarrow F=\left( 15 \right)\left( 7 \right) \\\ & \Rightarrow F=105.....(1) \\\ \end{aligned}$$ Now we have to choose 2 numbers among 50 numbers. So, let us assume the number of ways to pick 2 numbers among 50 numbers is equal to T. Here T is equal to the sum of a number of ways to pick one prime number among 15 prime numbers and to pick another number from the remaining 35 numbers and a number of ways to pick 2 numbers from the remaining numbers. $$\begin{aligned} & \Rightarrow T=\left( ^{15}{{C}_{1}}^{35}{{C}_{1}} \right){{+}^{35}}{{C}_{2}}{{+}^{15}}{{C}_{2}} \\\ & \Rightarrow T=\left( \dfrac{15!}{1!\left( 15-1 \right)!} \right)\left( \dfrac{35!}{1!\left( 35-1 \right)!} \right)+\dfrac{35!}{2!(35-2)!}+\dfrac{15!}{2!\left( 15-2 \right)!} \\\ & \Rightarrow T=\left( \dfrac{15!}{1!14!} \right)\left( \dfrac{35!}{1!34!} \right)+\dfrac{35!}{2!33!}+\dfrac{15!}{2!13!} \\\ & \Rightarrow T=\left( \dfrac{\left( 15 \right)\left( 14! \right)}{1!14!} \right)\left( \dfrac{\left( 35 \right)\left( 34! \right)}{1!34!} \right)+\dfrac{\left( 35 \right)\left( 34 \right)\left( 33! \right)}{2!33!}+\dfrac{\left( 15 \right)\left( 14 \right)\left( 13! \right)}{2!13!} \\\ & \Rightarrow T=(15)(35)+\dfrac{(35)(34)}{2}+\dfrac{(15)(14)}{2} \\\ & \Rightarrow T=525+595+105 \\\ & \Rightarrow T=1225....(2) \\\ \end{aligned}$$ Let us assume the probability to pick 2 prime numbers among 50 prime numbers is equal to P. $$\Rightarrow P=\dfrac{F}{T}.....(3)$$ Let us substitute equation (1) and equation (2) in equation (3), then we get $$\begin{aligned} & \Rightarrow P=\dfrac{105}{1225} \\\ & \Rightarrow P=\dfrac{15}{175} \\\ & \Rightarrow P=\dfrac{3}{35}.....(3) \\\ \end{aligned}$$ The probability that both the tickets drawn have prime numbers is equal to $$\dfrac{3}{35}$$. **Hence, option D is correct.** **Note:** Students may have confusion while finding the total number of outcomes. They may forget to include the number of ways to pick 2 prime numbers from 15 prime numbers in the total number of outcomes. They may forget to include the number of ways to pick one number which is prime and another number which is not a prime the total number of outcomes. In this way, many misconceptions can be made by students which results in the wrong answer. Hence, these misconceptions should get avoided.