College Math - Probability Word Problem

Question

Results from a survey shows that 63% of adults want access to their cellphones and laptops while on vacation. Selecting 8 adults at random determine the probability that exactly 3 of them want access to a cellphone.

 

This is a question from a quiz I already submitted but received the wrong solution too. I'm entirely stumped as to how the correct answer would be .09710. Thanks for any help or insight. 

 

 

Answer 1

Hi Dan,

 

Here's one way that I thought about it.

 

In the 8 adults, say that the first 3 want access and the last 5 don't. The probability of that is: (.63)^3 * (1-.63)^(8-3)

 

However, there are 8C3 (8 choose 3) variations of the scenario mentioned, so it needs to be multiplied by 8C3 = 8!/3!/5! = 8*7 = 56.

 

That gives:

 

(.63)^3 * (1-.63)^(8-3) * 8C3 = 0.09710

 

Hope this helps!