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Rachel has 10 sweets in a bag.

7 of the sweets are yellow.

3 of the sweets are red.

Rachel takes random sweets from the bag, one at a time, until she gets a red sweet.

She does not replace the sweets that she takes from the bag.

Rachel stops as soon as she gets a red sweet.

Work out the probability that Rachel takes no more than three sweets from the bag.


Sagot :

Using it's concept, it is found that there is a 0.713 = 71.3% probability that Rachel takes no more than three sweets from the bag.

What is a probability?

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that:

  • One sweet, the probability is 3/10.
  • For two, the probability is 7/10 then 3/9.
  • For three, it is 7/10 then 6/9 then 3/8.

So:

[tex]p = \frac{3}{10} + \frac{7}{10} \times \frac{3}{9} + \frac{7}{10} \times \frac{6}{9} \times \frac{3}{8} = 0.713[/tex]

0.713 = 71.3% probability that Rachel takes no more than three sweets from the bag.

More can be learned about the probabilities at https://brainly.com/question/24826394

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