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Thanks in advance for the help

 Jun 20, 2022
 #1
avatar+2222 
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a) \({20 \choose 4} = 4845 \) ways to choose the 4 members

 

There are 3 ways to choose which "Josh" is in the delegation. 

 

There is also \({17 \choose 3} = 680\) ways to choose the remaining 3 people (The 3 people named "Josh" are off-limits)

 

This makes for a probability of \({680 \times 3 \over 4845} \approx {\color{brown}\boxed{0.421}}\)

 Jun 21, 2022
 #2
avatar+2222 
0

b) There are \({20 \choose 4} =4845\) ways to choose the 4 students. 

 

However, the 3 people named "Josh" are off-limits, so there are only 17 people to choose the delegation from. 

 

This makes for \({17 \choose 4} = 2380\)

 

Thus, the total probability is \({2380 \over 4845} \approx \color{brown}\boxed{0.491}\)

 Jun 21, 2022
 #3
avatar+2222 
+1

c) The total probability of at least one "Josh" is the 1 - the probability of 0 "Josh" (what we don't want). 

 

This is called complementary counting, and it is based on the idea that if you remove what you don't want, you are left with the probability that you succeed. 

 

We already know from b that the probability of not getting a single "Josh" is \(0.491\), so the probability of at least one "Josh" is \(1 - 0.491 = \color{brown}\boxed{0.509}\)

 Jun 21, 2022

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