Question

From a group of 15 chess players, 8 are selected by lot to represent the group at a convention. What is the probability that the 8 selected include 3 of the 4 best players in the group.

Answer 1

Answer:

0.287

Step-by-step explanation:

Including 3 of the 4 best can be done in 4 ways

Remaining 5 have to be selected from 11...11C5 ways i.e. 462 ways

Total number of ways to pick 8 from 15 is 15C8 i.e. 6435

So probability would be 4*462/6435 =0.287

Answer 2

Given:

Total number of chess players = 15

Number of best payers = 4

Number of players to be selected for the convention = 8

To find:

Probability of selecting 3 out of 4 best payers in the group.

Solution:

Total Number of players = 15

Number of best payers = 4

Step 1 of 5

Number of ways 3 best players can be selected from 4 players = 4C3

= $\frac{4!}{3!(4-3)!}$

=4

Step 2 of 5

Now, the remaining 5 players will be selected from the remaining 11 players

Number of ways 5 players can be selected out of 11 players = 11C5

=$\frac{11!}{5!(11-5)!}$

=462

Step 3 of 5

Number of ways 3 best players and 5 normal players can be selected in the group = 462 x 4 = 1848

Step 4 of 5

Now, the number of ways 8 players can be selected out of 15 players = 15C8

= $\frac{15!}{8!(15-8)!}$

= 6435

Step 5 of 5

The probability that the 8 selected includes 3 of the 4 best players in the group = $\frac{Number of ways 3 best players can be selected in the group }{total number of ways of selecting 8 players in the group}$

= $\frac{1848}{6435}$

=0.287

The probability that the 8 selected includes 3 of the 4 best players in the group is 0.287