Final answer:
The probability of the sequences WWRWR and RWWWR from an urn containing two red and four white balls is calculated by multiplying the individual probabilities of each draw, since each draw is independent and with replacement.
Step-by-step explanation:
The subject of the question is probability, where we are asked to calculate the probability of obtaining two specific sequences of balls when drawing from an urn with replacement. The urn contains two red balls and four white balls, and the sequences given are WWRWR and RWWWR, each drawn in five trials.
To compute the probability of each sequence, we consider that each draw is independent. For instance, the probability of drawing a white ball, denoted as P(W), is ⅔ (four white balls out of six total balls). Similarly, the probability of drawing a red ball, P(R), is ⅓ (two red balls out of six total). As the draws are with replacement, these probabilities remain constant through all trials.
The probability of the sequence WWRWR is calculated as:
P(WWRWR) = P(W) * P(W) * P(R) * P(W) * P(R)
= ⅔ * ⅔ * ⅓ * ⅔ * ⅓
= (4/6) * (4/6) * (2/6) * (4/6) * (2/6)
P(WWRWR) is therefore the product of these probabilities.
Similarly, the probability of the sequence RWWWR is:
P(RWWWR) = P(R) * P(W) * P(W) * P(W) * P(R)
= ⅓ * ⅔ * ⅔ * ⅔ * ⅓
= (2/6) * (4/6) * (4/6) * (4/6) * (2/6)
P(RWWWR) is the product of these probabilities as well.
The requested probabilities of the given sequences are:
- The probability of the sequence WWRWR is (4/6)^3 * (2/6)^2.
- The probability of the sequence RWWWR is (2/6)^2 * (4/6)^3.
You can simplify the fractions and calculate the exact probabilities. Please mentioned the correct option in the final answer, as these are your individual probabilities for each sequence provided.