Question #351048

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.


Expert's answer

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

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

hence P(Tan)=10.30.20.20.10.1=0.1.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.P(Brown)+P(Red)=0.3+0.2=0.5.

ii. P(Ω)P(Yellow)=10.2=0.8.P(\Omega)-P(Yellow)=1-0.2=0.8.

iii.

P(Ω)P(Orange or Tan)=1P(Orange)P(Tan)=10.10.1=0.8.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(Ω)=1.P( brown\ or\ red\ or\ yellow\ or\ green\ or\ orange\ or\ tan)=P(\Omega)=1.


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