Question

As a special promotion for its 20-ounce bottles of soda, a soft drink company printed a message on the inside of each cap. Some of the caps said, “Please try again,” while others said, “You’re a winner!” The company advertised the promotion with the slogan “1 in 6 wins a prize.” Suppose the company is telling the truth and that every 20-ounce bottle of soda it fills has a 1-in-6 chance of being a winner. Seven friends each buy one 20-ounce bottle of the soda at a local convenience store. Let X = the number who win a prize. The probability distribution of X is shown here.

a)The store clerk is surprised when 3 of the friends win a prize. Is this group of friends just lucky, or is the company’s 1-in-6 claim inaccurate? Find the probability that at least 3 of a group of 7 friends would win a prize and use the result to justify your answer.

Winners	0	1	2	3	4	5	6	7
Probability	0.2791	0.3907	0.2344	0.0781	0.0156	0.0019	0.0001	0.000004

Answer 1

The given distribution of X is :

Winners	0	1	2	3	4	5	6	7
Probability	0.2791	0.3907	0.2344	0.0781	0.0156	0.0019	0.0001	0.000004

So to test the companies claim we will find the expected number of winners in a group of 7 people:

expected number of winners=  

hence the expected numbers of winners in a group of 7 people is just 1.166 that is approx 1 person.

so it's just that the friends are lucky enough to have won 3 out of 7, and the company's claim is correct because out of 7 it is expected that only 1.166 win so out of 6 it would be approximately equal to 1.

Now to find probability that atleast 3 out of 7 friends will win is

this is nothing but the sum of probabilities of 3,4,5,6,7 friends winning out of a group of 7 friends.

So above probability is the probability of atleast 3 friends winning from group of 7 friends.

As this probability is extremely small we can conclude that the chance that atleast 3 out of 7 friends will win the game is quite low.

Hence it is mere luck that 3 out of 7 have won it.