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Binomial distribution. A bowl contains 4 red balls, 3 blue balls and 6 black balls(Balls are put back after each pick)?
A: What is the probability of getting a red ball and after that a black ball ?
B: What is the probability of getting 2 red balls if you randomly chose 3 balls ?
C: Picking 4 balls, make a table that shows the probabilities of getting 0,1,2,3 or 4 red balls
D: Whats the probability of getting 3 or more, blue balls ?
2 réponses
- cidyahLv 7il y a 3 ans
There are 13 balls in the bowl.
P(red) = 4/13
P(blue )=3/13
P(black)=6/13
A.
P(red,black) = (4/13)(6/13) = 24/169
B.
n=3 (number of balls drawn)
p = 4/13 (probability of drawing a red ball)
x = number of red balls out of 3 draws
P(x=2) = 3C2 (4/13)^2 (1-4/13)^(3-2)
P(x=2) = (3) (4/13)^2 (9/13)^1
= 0.1966
C.
n=4
p=4/13
x=0,1,2,3,4
P(x=k) = 3Ck (4/13)^k (9/13)^(4-k)
0 0.229722
1 0.408390
2 0.272257
3 0.080668
4 0.008963
D.
You need to specify the number of times balls are drawn.
- Anonymeil y a 3 ans
b) Probability of different is probability of red on first (3/7) times probability of blue given a red has been chosen (4/6), plus the probability of blue on first (4/7) times probability of red given a blue has been chosen (3/6), or: 4/7.
