Answer: 0.0490
Step-by-step explanation:
Given : Number of blue balls = 6
Number of yellow balls = 7
Total number of balls = 13
Total number of ways for selecting 4 balls from 13 balls :-
[tex]^{13}P_4=\dfrac{13!}{(13-4)!}=17160[/tex]
The number of ways of selecting 3 blue and one yellow :-
[tex]^{6}P_3\times^{7}P_1=\dfrac{6!}{3!}\times\dfrac{7!}{6!}=840[/tex]
Now, the probability that of the balls selected, three of them will be blue and one of them will be yellow :-
[tex]\dfrac{840}{17160}\approx0.0490[/tex]
Hence, the required probability is 0.0490 .
