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Question
a box contains 8 letters, shown as reaching. what is the probability of the outcome hang in that order if 4 letters are drawn one by one (a) with replacement? (b) without replacement?
the probability of the outcome hang in that order if 4 letters are drawn with replacement is
(type your answer as a fraction.)
the probability of the outcome hang in that order if 4 letters are drawn without replacement is
(type your answer as a fraction.)
Step1: Analyze the letters in "REACHING"
The word "REACHING" has 8 letters: R, E, A, C, H, I, N, G. We need to find the probability of drawing H, A, N, G in order.
Part (a): With replacement
Step2: Probability of each letter with replacement
- Probability of drawing H first: There is 1 H out of 8 letters, so \( P(H) = \frac{1}{8} \).
- Probability of drawing A next: There is 1 A out of 8 letters (since we replace), so \( P(A) = \frac{1}{8} \).
- Probability of drawing N next: There is 1 N out of 8 letters, so \( P(N) = \frac{1}{8} \).
- Probability of drawing G next: There is 1 G out of 8 letters, so \( P(G) = \frac{1}{8} \).
Step3: Multiply the probabilities (independent events)
Since the draws are with replacement, the events are independent. So the probability of HANG in order is \( P = \frac{1}{8} \times \frac{1}{8} \times \frac{1}{8} \times \frac{1}{8} = \frac{1}{4096} \).
Part (b): Without replacement
Step4: Probability of each letter without replacement
- Probability of drawing H first: 1 H out of 8 letters, \( P(H) = \frac{1}{8} \).
- Probability of drawing A next: Now there are 7 letters left, 1 A, so \( P(A) = \frac{1}{7} \).
- Probability of drawing N next: Now 6 letters left, 1 N, so \( P(N) = \frac{1}{6} \).
- Probability of drawing G next: Now 5 letters left, 1 G, so \( P(G) = \frac{1}{5} \).
Step5: Multiply the probabilities (dependent events)
The probability of HANG in order without replacement is \( P = \frac{1}{8} \times \frac{1}{7} \times \frac{1}{6} \times \frac{1}{5} = \frac{1}{1680} \).
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(a) \(\frac{1}{4096}\)
(b) \(\frac{1}{1680}\)