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 = [tex]^{50}C_6[/tex]
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= [tex]^{44}C_{6}[/tex]
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 [tex]\dfrac{^{44}C_{6}}{^{50}C_6}[/tex]
[tex]=\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[/tex]
Hence, the required probability = 0.444225
