Given a spinner with four sections of equal size labeled A, B, C, and D, what is the probability of NOT getting an A after spinning the spinner two times? A. 15/16 B. 9/16 C. 1/2 D. 1/4 E. 1/8 The OA is B. How can I calculate this probability? Experts, can you give me an explanation here? Thanks.

Given a spinner with four sections of equal size labeled A, B, C, and D, what is the probability of NOT getting an A after spinning the spinner two times? A. 15/16 B. 9/16 C. 1/2 D. 1/4 E. 1/8 The OA is B. How can I calculate this probability? Experts, can you give me an explanation here? Thanks. Hi M7MBA, Let's take a look at your question. Probability of Not Getting an A after spinning the spinner two times can be calculated as: $$P\left(\text{Not getting an A}\right)\times P\left(\text{Not getting an A}\right)$$ $$=\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}$$ Therefore, option B is correct. Hope it helps. I am available if you'd like any follow up.

M7MBA wrote:Given a spinner with four sections of equal size labeled A, B, C, and D, what is the probability of NOT getting an A after spinning the spinner two times? A. 15/16 B. 9/16 C. 1/2 D. 1/4 E. 1/8 Since the 4 sections each have EQUAL sizes, P(getting A on a single spin) = 1/4 So, P(NOT getting A on a single spin) = 1 - 1/4 = 3/4 What is the probability of NOT getting an A after spinning the spinner two times? P(neither spin lands on A) = P(no A on 1st spin AND no A on 2nd spin) = P(no A on 1st spin) x P(no A on 2nd spin) = 3/4 x 3/4 = 9/16 = B Cheers, Brent

M7MBA wrote:Given a spinner with four sections of equal size labeled A, B, C, and D, what is the probability of NOT getting an A after spinning the spinner two times? A. 15/16 B. 9/16 C. 1/2 D. 1/4 E. 1/8 The OA is B. How can I calculate this probability? Experts, can you give me an explanation here? Thanks. The probability of not getting an A each spin is ¾. Therefore, the probability of not getting an A twice in two spins is ¾ x ¾ = 9/16. Answer: B

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