Question

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 1

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;

$= \frac{^2C_1 \times ^4C_1}{^6C_2} \\\\= \frac{8}{15}\\\\= 0.53333$

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

Answer 2

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

$6C2=\frac{6!}{(6-2)!2!} \\\\6C2=\frac{6!}{4! \times 2!} \\\\6C2 = 15$

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

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

$2C1= 2$

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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