Question

A bag contains 7 red marbles and 4 blue marbles. A marble is taken at random from the bag and replaced. Another marble is taken from the bag. Work out the probability that the two marbles taken from the bag are the same colour. Give your answer as a fraction.

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Answer 1

Answer:

$\sf \dfrac{65}{121}$

Step-by-step explanation:

Probability is a measure of how likely events are to happen.

When all the possible outcomes are equally likely, we can use this formula to calculate the probability of an event happening:

$\sf P(event)=\dfrac{\textsf{Number of outcomes where event happens}}{\textsf{Total number of possible outcomes}}$

Given that the bag contains 7 red marbles and 4 blue marbles, the total number of possible outcomes is 11.

Therefore:

$\textsf{Probability of drawing a red marble}=\sf \dfrac{7}{11}$

$\textsf{Probability of drawing a blue marble}=\sf \dfrac{4}{11}$

Given that the marble is replaced after it has been chosen, the probability of drawing two of each color are:

$\sf P(Red)\;and\;P(Red)=\dfrac{7}{11} \times \dfrac{7}{11}=\dfrac{49}{121}$

$\sf P(blue)\;and\;P(blue)=\dfrac{4}{11} \times \dfrac{4}{11}=\dfrac{16}{121}$

Therefore, the probability of drawing two reds OR two blues is:

$\sf P(Red+Red)\;or\;P(Blue+Blue)=\dfrac{49}{121}+\dfrac{16}{121}=\dfrac{65}{121}$

Answer 2

Answer:

semsee45 is right