Mr. Starnes and his wife have 6 grandchildren: Connor, Declan, Lucas, Piper, Sedona, and Zayne. They have 2 extra tickets to a holiday show, and will randomly select which 2 grandkids get to see the show with them. Find the probability that at least one of the girls (Piper and Sedona) get to go to the show.

Answer:

the probability that at least one of the girls (Piper and Sedona) get to go to the show is 0.533

Step-by-step explanation:

The computation of the probability that have at least one of the girl get to go is shown below;

[tex]= \frac{^2C_1 \times ^4C_1}{^6C_2} \\\\= \frac{8}{15}\\\\= 0.53333[/tex]

Hence, the probability that at least one of the girls (Piper and Sedona) get to go to the show is 0.533

So the same is relevant

Probability that at least one of the girls (Piper and Sedona) get to go to the show = 1/5

The number of grandchildren = 6

The number of girls = 2

Two grandkids out of 6 are to be selected

Number of ways of choosing 2 grandkids out of 6 is 6C2

[tex]6C2=\frac{6!}{(6-2)!2!} \\\\6C2=\frac{6!}{4! \times 2!} \\\\6C2 = 15[/tex]

Number of ways of selecting 2 grandkids out of 6 = 15 ways

Number of ways  of selecting 1 out of the two girls =2C1

[tex]2C1= 2[/tex]

Number of ways of selecting 1 out of the two girls = 2 ways

Number of ways of selecting of selecting both girls =2C2

Number of ways of selecting of selecting both girls = 1 way

Probability of selecting 1 out of the two grandkids = 1/15

Probability of selecting both girls = 2/15

Probability that at least one of the girls (Piper and Sedona) get to go to the show = 1/15 + 2/15

probability that at least one of the girls (Piper and Sedona) get to go to the show = 3/15

Probability that at least one of the girls (Piper and Sedona) get to go to the show = 1/5

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