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fishbate · Oct 14, 2011, 07:47 AM

An internet company has season tickets to the lakers basketball games. The company president always invites one of the 6 vice presidents to attend the games with him, and claims he selects the person to attend at random. One of the 6 vice presidents has not been invited to attend any of the last 4 lakers games. What is the likelihood this could be due to chance?

ebaines · Oct 14, 2011, 07:50 AM

The chance if any one VP being invited to a particular game is 1/6. Conversely, the chance of not being invited is 1 - 1/6 = 5/6. Given this, what is probability of not getting invited 4 times in a row? Post back with your answer and we'll check it for you.

fishbate · Oct 14, 2011, 07:59 AM

My answer was the following:

P(A/B)=6/6

(1/6)(2/5)(3/4(4/3)(5/2)(6/1)=0.9945

4-0.9945=3.0055 OR 5.5

please check.

ebaines · Oct 14, 2011, 08:13 AM

Sorry - not right. I don't understand what you're trying to do - what does P(A/B) mean? And why do you say that 3.0055 = 5.5? The answer to this problem is a probability value between 0 and 1, so if you get an answer > 1 you know it's not right.

Here's another hint: the probability of not being chosen for the first game is 5/6, right? The probability of not being chosen for the second game is also 5/6. So the probability of not being chosen for both the first and second games is 5/6 x 5/6 = 25/36. Do you see why that is? So now continue on - what's the proability of not being chose for any of the first 4 games?