Answer to Question #351048 in Statistics and Probability for Bob

Question

4. QUESTION:

M&M sweets are of varying colours and the different colours occur in different proportions. The table below gives the probability that a randomly chosen M&M has each colour, but the value for tan candies is missing.

Colour Brown Red Yellow Green Orange Tan

Probability 0.3 0.2 0.2 0.1 0.1 ?

(a) What value must the missing probability be?

(b) You draw an M&M at random from a packet. What is the probability of each of the following events?

i. You get a brown one or a red one.

ii. You don’t get a yellow one.

iii. You don’t get either an orange one or a tan one.

iv. You get one that is brown or red or yellow or green or orange or tan.

Answer 1

(a) Since the probability of the set of all possible outcomes $P(\Omega)=1,$ then

$P(Brown)+P(Red)+P(Yellow)+P(Green)\\+P(Orange)+P(Tan)=P(\Omega)=1$

hence $P(Tan)=1-0.3-0.2-0.2-0.1-0.1=0.1.$

(b)

i. $P(Brown)+P(Red)=0.3+0.2=0.5.$

ii. $P(\Omega)-P(Yellow)=1-0.2=0.8.$

iii.

$P(\Omega)-P(Orange\ or\ Tan)=1-P(Orange)-P(Tan)=1-0.1-0.1=0.8.$

iv. $P( brown\ or\ red\ or\ yellow\ or\ green\ or\ orange\ or\ tan)=P(\Omega)=1.$