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Suppose an opaque jar contains 3 red marbles and 10 green marbles. The following exercise refers to the experiment of picking two marbles from the jar without replacing the first one. What is the probability of getting a green marble and a red marble? (Enter your probability as a fraction. Hint: How is this exercise different from finding the probability of getting a green marble first and a red marble second?)

Respuesta :

Answer:

[tex]\frac{5}{13}[/tex]

Step-by-step explanation:

Given,

Red marbles = 3,

Green marbles = 10,

So, the total marbles = 3 + 10 = 13,

[tex]\because \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]

Since, here replacement is not allowed,

Thus, the probability of getting a green marble and a red marble

= first red and second green + first green second red

[tex]=\frac{3}{13}\times \frac{10}{12}+\frac{10}{13}\times \frac{3}{12}[/tex]

[tex]=2\times \frac{3}{13}\times \frac{10}{12}[/tex]

[tex]=\frac{30}{78}[/tex]

[tex]=\frac{5}{13}[/tex]

Note : The probability of getting a green marble first and a red marble second

= [tex]\frac{10}{13}\times \frac{3}{12}[/tex]

[tex]=\frac{30}{156}[/tex]

[tex]=\frac{5}{26}[/tex]

The probability of getting a green marble and a red marble when picking two marbles from the jar without replacing the first one is;  ⁵/₁₃

What is the selection probability?

We are given;

Number of red marbles = 3

Number of green marbles = 10,

Thus,

Total marbles = 3 + 10

Total marbles = 13

Probability of first red and second green is;

P(Red then Green) = (³/₁₃ * ¹⁰/₁₂)

Probability of first green and second red is;

P(green then red) = (¹⁰/₁₃ * ³/₁₂)

Thus,  probability of getting a green marble and a red marble is;

P(Green and red) = (³/₁₃ * ¹⁰/₁₂)  + (¹⁰/₁₃ * ³/₁₂)

P(Green and red) = ⁵/₁₃

Read more on probability selection at; https://brainly.com/question/251701

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