Question

# Recall the pre-lab candy situation consider: two unopened bags of candy (one Skittles the other M&M's)....

Recall the pre-lab candy situation consider: two unopened bags of candy (one Skittles the other M&M's). The manufacturers state the distribution of coloured candies is:

M&M'S MILK CHOCOLATE: 24% blue, 19% orange, 16% green, 13% yellow, 13% red, 15% brown.

SKITTLES: colours are reportedly distributed evenly, meaning each colour (red, orange, green, blue, yellow, and purple) has a probability of 1616.

Assume that both of these unopened bags have the same number of candies. If they are opened and mixed, and one candy is randomly selected; what is the probability it is a Skittle? [Hint: only focus on the colour you need]

?(Skittle | blue)= ????

P(Skittle) = 0.5

P(M&M) = 0.5

P(blue | Skittle) = 0.16

P(blue | M&M) = 0.24

P(blue) = P(blue | Skittle) * P(Skittle) + P(blue | M&M) * P(M&M)

= 0.16 * 0.5 + 0.24 * 0.5

= 0.2

P(Skittle | blue) = P(blue | Skittle) * P(Skittle) * P(blue) = 0.16 * 0.5 / 0.2 = 0.4 (ans)

Probability that the candy selected is a skittle = 0.5

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