A bag contains 5 red balls and 3 green balls. If two balls are drawn at random without replacement, what is the probability that both balls drawn are red?
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When calculating probabilities in dependent events (like drawing without replacement), multiply the probabilities of each step. Always adjust the total number of outcomes after each draw.
A bag contains 5 red balls and 3 green balls. We need to calculate the probability of drawing two red balls without replacement. Step 1: Total number of balls Total balls in the bag: \[ 5 \, \text{(red balls)} + 3 \, \text{(green balls)} = 8 \, \text{balls} \] Step 2: Probability of drawing the first red ball Probability of drawing the first red ball: \[ P(\text{First red}) = \frac{5}{8} \] Step 3: Probability of drawing the second red ball After drawing one red ball, 4 red balls and a total of 7 balls remain. Probability of drawing the second red ball: \[ P(\text{Second red}) = \frac{4}{7} \] Step 4: Multiply the probabilities For dependent events (drawing without replacement), the probability of both balls being red is the product of individual probabilities: \[ P(\text{Both red}) = P(\text{First red}) \times P(\text{Second red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{21} \] Answer: The probability of drawing two red balls is \( \frac{5}{21} \).