Question

Xander spends most of his time with his 10 closest friends. he has known 4 of his 10 friends since kindergarten. if he is going to see a movie tonight with 3 of his 10 closest friends, what is the probability that the first 2 of the friends to show up to the movie are friends he has known since kindergarten but the third is not? startfraction 1 over 30 endfraction startfraction 9 over 125 endfraction startfraction 12 over 125 endfraction startfraction 1 over 10 endfraction

Answer 1

Probability that the first 2 of the friends to show up to the movie are friends he has known since kindergarten but the third not is 1/10

Total friends = 10

Friends from kindergarten = 4

Probability is the chance that a given event will occur. Probability of an event lies within 0 to 1

P(E) = Favourable outcomes / Total outcomes

Probability of getting 1st friend from kindergarten = 4/10

Probability of getting 2nd friend from kindergarten = 3/9

Probability of getting 3rd friend not from kindergarten = 6/8

Since all these probabilities are independent, We can use Multiplicative identity.  Thus,

Required probability is 4/10 * 3/9 * 6/8

= 1/10

Thus, Probability that the first 2 of the friends to show up to the movie are friends he has known since kindergarten but the third not is 1/10

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