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Albert tosses 5 biased coins with a probability of heads as 0.6
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Updated on: 14 Oct 2022, 21:53
Updated on: 14 Oct 2022, 21:53
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Albert tosses 5 biased coins with a probability of heads as 0.6 and the probability of tails as 0.4. What is the probability that 3 of these coins will land on heads?
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\(10(0.6)^3(0.4)^2\)
Originally posted by Anna7 on 22 Sep 2022, 23:40.
Last edited by BrushMyQuant on 14 Oct 2022, 21:53, edited 5 times in total.
Last edited by BrushMyQuant on 14 Oct 2022, 21:53, edited 5 times in total.
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Re: Albert tosses 5 biased coins with a probability of heads as 0.6
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23 Sep 2022, 02:03
23 Sep 2022, 02:03
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I do not see any option as answer choices or this is a numeric entry question?
please follow our rules for posting https://gre.myprepclub.com/forum/rules- ... -1083.html
I do not see any option as answer choices or this is a numeric entry question?
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Re: Albert tosses 5 biased coins with a probability of heads as 0.6
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14 Oct 2022, 21:52
14 Oct 2022, 21:52
1
Given that Albert tosses 5 biased coins with a probability of heads as 0.6 and the probability of tails as 0.4. We need to find What is the probability that 3 of these coins will land on heads?
5 Biased coins are tosses => Number of cases = \(2^5\) = 32
Now out of the five place _ _ _ _ _ we need to find three places where Heads can come => 5C3 ways = \(\frac{5!}{3!*(5-3)!}\) ways = \(\frac{5*4*3!}{3!*2!}\) ways = 10 ways
=> P(3H) = P(Three Heads and 2 Tails) = Number of places * P(Head) * P(Head) * P(Head) * P(Tail) * P(Tail) = 10*0.6*0.6*0.6*0.4*0.4 = \(10(0.6)^3(0.4)^2\)
So, Answer will be \(10(0.6)^3(0.4)^2\)
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
5 Biased coins are tosses => Number of cases = \(2^5\) = 32
Now out of the five place _ _ _ _ _ we need to find three places where Heads can come => 5C3 ways = \(\frac{5!}{3!*(5-3)!}\) ways = \(\frac{5*4*3!}{3!*2!}\) ways = 10 ways
=> P(3H) = P(Three Heads and 2 Tails) = Number of places * P(Head) * P(Head) * P(Head) * P(Tail) * P(Tail) = 10*0.6*0.6*0.6*0.4*0.4 = \(10(0.6)^3(0.4)^2\)
So, Answer will be \(10(0.6)^3(0.4)^2\)
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
