Question

In a​ lottery, 6 numbers are randomly sampled without replacement from the integers 1 to 50. Their order of selection is not important. Find the probability of holding a ticket that has zero winning numbers out of the 6 numbers selected for the winning ticket out of the 50 possible numbers.

Answer 1

Answer: 0.444225

Step-by-step explanation:

Given : The total number of tickets = 50

Number of tickets are randomly sampled without replacement  =6

Since the order of selection is not important , so we use combinations.

Total number of ways to select 6 tickets = $^{50}C_6$

The number of winning tickets = 6

So, number of tickets that are not winning = 50-6=44

Number of ways of selecting zero winning numbers= $^{44}C_{6}$

Now , the probability of holding a ticket that has zero winning numbers out of the 6 numbers selected for the winning ticket out of the 50 possible numbers would be $\dfrac{^{44}C_{6}}{^{50}C_6}$

$=\dfrac{\dfrac{44!}{6!(44-6)!}}{\dfrac{50!}{6!(50-6)!}}\\\\\\=\dfrac{\dfrac{44\times43\times42\times41\times40\times39\times38!}{38!}}{\dfrac{50\times49\times48\times47\times46\times45\times44!}{44!}}\\\\\\=\dfrac{44\times43\times42\times41\times40\times39}{50\times49\times48\times47\times46\times45}\\\\=\dfrac{252109}{567525}\approx0.444225$

Hence, the required probability = 0.444225