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A bag contains 3 red and 5 green sweets. Tim takes a sweet at random and eats it. He then takes another sweet. What is the probability that tim takes 2 red sweets

Respuesta :

JeanaShupp

Answer: [tex]\dfrac{3}{28}[/tex].

Step-by-step explanation:

Given : A bag contains 3 red and 5 green sweets. Tim takes a sweet at random and eats it.

Total sweets were in bag = 5+3=8

Number of ways to select 2 red sweets = [tex]3\times2=6[/tex]

Number of ways to select any two sweets = [tex]8\times7=56[/tex]

Now, the probability that Tim takes 2 red sweets is given by :-

[tex]=\dfrac{\text{Number of ways to take red sweets}}{\text{Number of ways to select any two sweets}}\\\\=\dfrac{6}{56}=\dfrac{3}{28}[/tex]

Hence, the probability that Tim takes 2 red sweets is [tex]\dfrac{3}{28}[/tex].

fichoh

The probability that Tim take 2 red Sweet without replacement is 0.0476

Given the Parameters :

  • Red Sweets = 2
  • Green sweets = 5
  • Total number of sweets = 2 + 5 = 7

Probability = Required outcome / Total possible outcomes

P(2 red sweets) = P(red) × P(red)

Since, selection is without replacement ;

P(2 red sweets) = 2/7 × 1/6 = 0.0476

Therefore, the probability of picking two red sweets is 0.0476

Learn more : https://brainly.com/question/18405415

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