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For one toss of a certain coin, the probability that the outcome is he
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09 Jul 2012, 06:50
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For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times? A. \((0.6)^5\) B. \(2(0.6)^4\) C. \(3(0.6)^4\) D. \(4(0.6)^4(0.4) + (0.6)^5\) E. \(5(0.6)^4(0.4) + (0.6)^5\)
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Re: For one toss of a certain coin, the probability that the outcome is he
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09 Jul 2012, 06:57



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Re: For one toss of a certain coin, the probability that the outcome is he
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19 Sep 2014, 19:00
Bunuel wrote: Stiv wrote: For one toss of a certain coin, the probability that the outcome is heads is o.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?
A. (0.6)^5 B. 2(0.6)^4 C. 3(0.6)^4 D. 4(0.6)^4(0.4) + (0.6)^5 E. 5(0.6)^4(0.4) + (0.6)^5 \(P(h)=0.6\), so \(P(t)=0.4\). We want to determine the probability of at least 4 heads in 5 tries. At least 4 heads means 4 or 5. Let's calculate each one: 5 heads: \(P(h=5)=0.6^5\); 4h and 1t: \(P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4\), multiplying by 5 as 4h and 1t may occur in 5 different ways: hhhht hhhth hhthh hthhh thhhh So, \(P(h\geq{4})=0.6^5+5*0.6^4*0.4\). Answer: E. Hi Bunuel, Thanks for the explanation. I am having difficulty understanding though when should we consider order to be important and when not? When it is not explicitly stated that is.. Please let me know.
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Re: For one toss of a certain coin, the probability that the outcome is he
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20 Sep 2014, 12:22
kpali wrote: Bunuel wrote: Stiv wrote: For one toss of a certain coin, the probability that the outcome is heads is o.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?
A. (0.6)^5 B. 2(0.6)^4 C. 3(0.6)^4 D. 4(0.6)^4(0.4) + (0.6)^5 E. 5(0.6)^4(0.4) + (0.6)^5 \(P(h)=0.6\), so \(P(t)=0.4\). We want to determine the probability of at least 4 heads in 5 tries. At least 4 heads means 4 or 5. Let's calculate each one: 5 heads: \(P(h=5)=0.6^5\); 4h and 1t: \(P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4\), multiplying by 5 as 4h and 1t may occur in 5 different ways: hhhht hhhth hhthh hthhh thhhh So, \(P(h\geq{4})=0.6^5+5*0.6^4*0.4\). Answer: E. Hi Bunuel, Thanks for the explanation. I am having difficulty understanding though when should we consider order to be important and when not? When it is not explicitly stated that is.. Please let me know. Think logically, the case of 4 heads and 1 tail can occur in several ways (each of which having the same probability), so we have to account for the order here.
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Re: For one toss of a certain coin, the probability that the outcome is he
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31 May 2017, 22:17
Stiv wrote: For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?
A. (0.6)^5 B. 2(0.6)^4 C. 3(0.6)^4 D. 4(0.6)^4(0.4) + (0.6)^5 E. 5(0.6)^4(0.4) + (0.6)^5 1. Probability of getting 4 heads and one tail is 5(0.6)^4 * (0.4). 5 is nothing but 5C4 ways. of 4 heads happening. 2. Probability of getting heads all the 5 times is (0.6)^5 3. Total probability is answer E.
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Re: For one toss of a certain coin, the probability that the outcome is he
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25 Jun 2017, 11:29
Stiv wrote: For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?
A. (0.6)^5 B. 2(0.6)^4 C. 3(0.6)^4 D. 4(0.6)^4(0.4) + (0.6)^5 E. 5(0.6)^4(0.4) + (0.6)^5 Refer to the solution in the picture
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Re: For one toss of a certain coin, the probability that the outcome is he
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18 Oct 2018, 17:13
Stiv wrote: For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?
A. \((0.6)^5\)
B. \(2(0.6)^4\)
C. \(3(0.6)^4\)
D. \(4(0.6)^4(0.4) + (0.6)^5\)
E. \(5(0.6)^4(0.4) + (0.6)^5\) The outcome “at least 4 heads out of 5 flips” means that we can get either exactly 4 heads out of 5 flips OR 5 heads out of 5 flips. First, we need to determine the probability of exactly 4 heads out of 5 flips, so: P(HHHHT) = (0.6)^4 x 0.4 Since HHHHT can be arranged in 5!/4! = 5 ways, the total probability of obtaining exactly 4 heads is 5(0.6)^4 x 0.4. The probability of 5 heads is (0.6)^5. So the total probability is 5(0.6)^4 x 0.4 + (0.6)^5. Answer: E
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