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Assume that there are 7 different issues of Time magazine, 8 different issues of Popular Science, and 4 different issues of Sports Illustrated, including the December 1st issue, on a rack. You choose 4 of them at random.

What is the probability that you choose at least 2 of the Popular Science magazines?

Respuesta :

Let the letters T, P, S represent Time, Popular Science and Sports Illustrated magazines respectively.

We have 7 T [tex]\{T_1, T_2, T_3, ...T_7\}[/tex], 8 P [tex]\{P_1, P_2, P_3, ...P_8\}[/tex] and 4 S  [tex]\{S_1, S_2, S_3, S_4\}[/tex].


The total number of ways of selecting 4 out of 7+8+4=19 is given by the formula:

[tex]C(19, 4)= \frac{19!}{4!15!}=\displaystyle{ \frac{19\cdot18\cdot17\cdot16\cdot15!}{4!15!}= \displaystyle{ \frac{19\cdot18\cdot17\cdot16}{4\cdot3\cdot2}=[/tex]

[tex]19\cdot6\cdot17\cdot2=3876[/tex]


The event " choosing at least 2 P" is the complement of the event "choosing at most 1 P"

so P(choosing at least 2 P)=1-P(choosing at most 1 P)

P(choosing at most 1 P) = P(choosing no P) + P(choosing exactly 1 P)


P(choosing no P) =  

[tex]\displaystyle{ \frac{C(11, 4)}{3876}= \frac{11!}{4!7!}\cdot\frac{1}{3876}= \frac{11*10*9*8}{4*3*2}\cdot\frac{1}{3876}= 330\cdot\frac{1}{3876}= 0.085[/tex]

because we are choosing all 4 from the 11 non P magazines


and 

P(choosing exactly 1 P) is 

[tex]\displaystyle{ \frac{8\cdot C(11,3)}{3876}=8\cdot \frac{11!}{3!8!} \cdot\frac{1}{3876}= \frac{11!}{3!7!} \cdot\frac{1}{3876}=4\cdot\frac{11!}{4!7!}\cdot\frac{1}{3876}[/tex]
[tex]=4\cdot0.085=0.34[/tex]

here we modified the operations of factorials, to make use of the previous calculation. 

Thus, 

P(choosing at most 1 P) = P(choosing no P) + P(choosing exactly 1 P)=0.085+0.34=0.425


So, P(choosing at least 2 P)=1-0.0425=0.575