Respuesta :
Answer:
The questions asked are
If you randomly select 4 peanuts
1. Compute the probability that exactly three of the four M&M’s are brown
2. Compute the probability that two or three of the four M&M’s are brown.
3. Compute the probability that at most three of the four M&M’s are brown.
4. Compute the probability that at least three of the four M&M’s are brown.
Step-by-step explanation:
Given the following information
Brown=12%. P(B)=0.12
Yellow=15%. P(Y)=0.15
Red=12%. P(R), =0.12
Blue=23%. P(B) =0.23
Orange, =23%. P(O) =0.23
Green=15%. P(G)=0.15
Question 1.
They are independent events
If there are exactly three brown and the last is not brown
P(B n B n B n B')
P(B)×P(B)×P(B)×P(B')
0.12×0.12×0.12×(1-P(B))
0.001728×(1-0.12)
0.001728×0.88
0.00152.
0.152%
2. If two or three are brown
I.e we are going to two brown and two none brown or three brown and one not brown. (P(B)×P(B)×P(B')×P(B'))+ (P(B)×P(B)×P(B'))
(0.12×0.12×0.88×0.88)+(0.12×0.12×0.12×0.88)
0.0112+0.00152
0.0127
1.27%
3. At most three brown out of four then we are going to have
BBBB', BBB'B', BB'B'B', B'B'B'B'
These are the cases of at most three brown.
P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B')×P(B') + P(B)×P(B')×P(B')×PB')+ P(B')×P(B')×P(B')×P(B')=
0.12×0.12×0.12×0.88+ 0.12×0.12×0.88×0.88+ 0.12×0.88×0.88×0.88+ 0.88×0.88×0.88×0.88=0.694
0.694
69.4%
4. At least 3 brown out of four selection
I.e BBBB', BBBB
These are the two options
P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B)×P(B)=
0.12×0.12×0.12×0.88 + 0.12×0.12×0.12×0.12
0.001728
0.173%
Answer:
The probability that the first green candy is the seventh M&M selected = 0.0566
Step-by-step explanation:
Probability of selecting a green candy = P(G) = 15% = 0.15
Probability of not selecting a green candy, P(G') = 1 - 0.15 = 0.85
To compute the probability that the first green candy is the seventh M&M selected
For this to happen, the not green candies are selected 6 times, before a green one is drawn on the 7th draw
That is
[P(G')]⁶ (P(G) = 0.85⁶ × 0.15 = 0.05657 = 0.0566
