Step-by-step explanation:
I assume it is meant to draw 2 blue balls in a row out of the full pool of balls (no other draws happen before or during this process).
tricky, your teacher.
I think the temptation is to simply say 1/3×1/3 = 1/9.
but is it ?
the probabilty to draw one blue ball out of the complete system (no ball missing) is
x / (6+4+x) = 1/3
x = (6+4+x)/3
3x = 6 + 4 + x = 10 + x
2x = 10
x = 5
so, we have a system of 6 orange, 4 green and 5 blue balls (15 all together) .
the probability at our first pull to draw a blue ball is the indicated 5/15 = 1/3.
but now, for the second ball, we have only 6 orange, 4 green and 4 blue balls in the pool (14 altogether).
and so the probability to draw a blue ball out of this is
4/14 = 2/7.
therefore, the probability to draw 2 blue balls is now the combination
1/3 × 2/7 = 2/21
aha ! so, it is a little bit less than the originally suspected probability of 1/9.
Answer:
answer a
Step-by-step explanation:
Answer:
Yes
Step-by-step explanation:
Permutation :
nPr = n! ÷ (n - r)!
nCr = n! ÷ (n-r)!r!
11 objects taken 3 at a time :
11P3 = 11! ÷ (11 - 3)!
11P3 = 11! / 8!
11P3 = (11 * 10 * 9 )
11P3 = 990
Dividing by 3!
3! = 3*2*1 = 6
= 990 / 6
= 165
For combination :
11C3 = 11! ÷ 8!3!
11C3 = (11 * 10 * 9) / (3 * 2 * 1)
11C3 = 990 / 6
11C3
