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Sagot :
Answer:
P(at least 1 infected person) = 0.7
Explanation:Note that:
Probability = (Number of possible outcomes) / (Number of total outcomes)
The crew members = 5
Number of infected crew members = 2
Number of uninfected crew members = 3
We want to select two crew members for a task out of 5
Number of total outcomes = 5C2 (Selecting 2 out of 5)
Note that:
[tex]nCr=\frac{n!}{(n-r)!r!}[/tex][tex]\begin{gathered} 5C2=\frac{5!}{(5-2)!2!} \\ \\ 5C2=\frac{5!}{3!2!} \\ \\ 5C2=\frac{5\times4\times\cancel{3!}}{\cancel{3!}\times(2\times1)} \\ \\ 5C2=\frac{20}{2} \\ \\ 5C2=10 \end{gathered}[/tex]Number of total outcomes = 10
That is, there are 10 ways of selecting 2 crew members from 5
Number of ways of selecting 1 infected person means how we can select two people out of 5 such that 1 one of them will be infected. This means that we will select 1 from the two infected persons, and select the second one from the 3 uninfected persons
Number of ways of selecting 1 infected persons = 2C1 x 3C1
[tex]\begin{gathered} 2C1=\frac{2!}{(2-1)!1!}=\frac{2!}{1!1!}=\frac{2\times1}{1\times1} \\ 2C1=2 \\ \\ 3C1=\frac{3!}{(3-1)!1!}=\frac{3!}{2!1!}=\frac{3\times2\times1}{2\times1\times1}=\frac{6}{2} \\ 3C1=3 \end{gathered}[/tex]Number of ways of selecting 1 infected persons = 2 x 3
Number of ways of selecting 1 infected persons = 6
Number of ways of selecting 2 infected persons = 2C2
[tex]2C2=\frac{2!}{(2-2)!2!}=\frac{2!}{2!}=1[/tex]Number of ways of selecting 2 infected persons = 1 way
Probability of selecting 1 infected person = 6/10 = 0.6
Probability of selecting 2 infected person = 1/10 = 0.1
P(at least 1 infected person) = P(1 infected person) + P(2 infected persons)
P(at least 1 infected person) = 0.6 + 0.1
P(at least 1 infected person) = 0.7
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