Respuesta :
a) P(BL') = 0.77
b) P(BL) + P(G) = 0.380
c) 0.012167 = 0.012 to 3 d.p
d) 0.456533 = 0.456 to 3 d.p
e) 0.543467= 0.543 to 3 d.p
Probability of brown M&M's = P(Br) = 0.12
Probability of yellow M&M's = P(Y) = 0.15
Probability of red M&M's = P(R) = 0.12
Probability of blue M&M's = P(BL) = 0.23
Probability of orange M&M's = P(O) = 0.23
Probability of green M&M's = P(G) = 0.15
Total probability = 1
a) The probability that a randomly selected peanut M&M is not blue = 1 - P(BL) = 1 - 0.23 = 0.77
b) Tblue he probability that a randomly selected peanut M&M is blue or green = P(BL) + P(G) = 0.23 + 0.15 = 0.38
c) The probability that three randomly selected peanut M&M's are all orange= P(O) × P(O) × P(O) = 0.23 × 0.23 × 0.23 = 0.012167
d) If you randomly select three peanut M&M's, compute that probability that neither of them are orange
This probability = P(O') × P(O')×P (O')
P(O') = 1 - P(0) = 1 - 0.23 = 0.77
P(NO)=P(O') × P(O') × P(O')= 0.77 × 0.77× 0.77 = 0.456533
e) If you randomly select three peanut M&M's, compute that probability that at least one of them is Orange
let the probability of at least 1 of them being orange be P(AO)
let the probability of none of them being orange be P(NO)
we know that
P(AO) = 1- P(NO)
= 1- 0.456533
= 0.543467
learn more about of probability here
https://brainly.com/question/14854781
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