You have a bag of 63 candies you want to give to your 9 friends. The company that makes the candies guarantees that exactly 9 of the candies in the bag are red, the most delicious color. Anyone who doesn't get a red candy will be so upset that they will stop being your friend! But the candies are in identical wrappings, so you are forced to give each friend 7 candies and hope for the best. What's the probability you lose one or more friends

the probability you lose one or more friends is 0.9836

Step-by-step explanation:

 Given the data in the question;

P( you lose one or more friend ) = 1 - P( keeping all friends )

Now, keeping all friends is possible only when each friend receives red candy.

As there are only 9 red candies and 9 friends, 1 red candy needs to be given to each friend out of 9 candies

Thus, total candies given = 63

and each friend will get 8 non red and 1 red candy

Thus, P( keeping all friends )

= [ ⁹C₁ × ⁹C₁ × ⁹C₁ × ⁹C₁ × ⁹C₁ × ⁹C₁ × ⁹C₁ × ⁹C₁ × ⁹C₁ ] / × ⁶³C₉

SO

⁹C₁  = 9!/(1!(9-1)!) = 9 and ⁴⁹C₇ = 63!/(9!(63-9)!)  = 23,667,689,815

so

P( keeping all friends ) = [ 9 × 9 × 9 × 9 × 9 × 9 × 9 × 9 × 9 ] / 23,667,689,815

P( keeping all friends ) = 387,420,489 / 23,667,689,815 = 0.016369

P( you lose one or more friend ) = 1 - P( keeping all friends )

P( you lose one or more friend ) = 1 - 0.016369

P( you lose one or more friend ) = 0.9836

Therefore, the probability you lose one or more friends is 0.9836

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?

according to egenuity the answer is

step-by-step explanation:

d. 344.22