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A coin is weighted so that the probability of heads on any flip is 0.6

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A coin is weighted so that the probability of heads on any flip is 0.6  [#permalink]

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New post 14 Sep 2015, 21:41
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A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?


A. \((0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2\)

B. \((0.6)^5 + 4(0.6)^4(0.4) + 6(0.6)^3(0.4)^2\)

C. \((0.6)^5 + 3(0.6)^4(0.4) + 2(0.6)^3(0.4)^2\)

D. \((0.6)^5 + 2(0.6)^4(0.4) + (0.6)^3(0.4)^2\)

E. \((0.6)^5 + (0.6)^4(0.4) + (0.6)^3(0.4)^2\)


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Re: A coin is weighted so that the probability of heads on any flip is 0.6  [#permalink]

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New post 14 Sep 2015, 22:32
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Bunuel wrote:
A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?

A. (0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2
B. (0.6)^5 + 4(0.6)^4(0.4) + 6(0.6)^3(0.4)^2
C. (0.6)^5 + 3(0.6)^4(0.4) + 2(0.6)^3(0.4)^2
D. (0.6)^5 + 2(0.6)^4(0.4) + (0.6)^3(0.4)^2
E. (0.6)^5 + (0.6)^4(0.4) + (0.6)^3(0.4)^2

Kudos for a correct solution.


Probability of Head, P(H) = 0.6
Probability of Tail, P(T) = 0.4

Tail will appear NO more than twice
i.e. favourable cases
2 Tails and 3 Heads, Probability = 5C2*(0.6)^3*(0.4)^2
1 Tail and 4 Heads, Probability = 5C1*(0.6)^4*(0.4)^2
0 Tail and 5 Heads, Probability = (0.6)^5

Required Probability = Sum of all Favourable cases = (0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2

Answer: option A
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Re: A coin is weighted so that the probability of heads on any flip is 0.6  [#permalink]

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New post 18 Sep 2015, 16:56
1
Required prob -

THHHH - 5!/(4!) (0.4)(0.6)^4

TTHHH - 5!/2!.3! (0.4)^2(0.6)^3


HHHHH-5!/5! (0.6)^5

The sum matches (A)
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Re: A coin is weighted so that the probability of heads on any flip is 0.6  [#permalink]

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New post 19 Sep 2015, 03:03
1
Answer is A
3 cases possible:
hhhhh or hhhht (5 cases) or hhhtt(10 cases)

Multiplying the respective probabilities with the weights gives us A
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Re: A coin is weighted so that the probability of heads on any flip is 0.6  [#permalink]

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New post 19 Sep 2015, 06:19
GMATinsight wrote:
Bunuel wrote:
A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?

A. (0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2
B. (0.6)^5 + 4(0.6)^4(0.4) + 6(0.6)^3(0.4)^2
C. (0.6)^5 + 3(0.6)^4(0.4) + 2(0.6)^3(0.4)^2
D. (0.6)^5 + 2(0.6)^4(0.4) + (0.6)^3(0.4)^2
E. (0.6)^5 + (0.6)^4(0.4) + (0.6)^3(0.4)^2

Kudos for a correct solution.


Probability of Head, P(H) = 0.6
Probability of Tail, P(T) = 0.4

Tail will appear NO more than twice
i.e. favourable cases
2 Tails and 3 Heads, Probability = 5C2*(0.6)^3*(0.4)^2
1 Tail and 4 Heads, Probability = 5C1*(0.6)^4*(0.4)^2
0 Tail and 5 Heads, Probability = (0.6)^5

Required Probability = Sum of all Favourable cases = (0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2

Answer: option A


Hi,
i think it is a typing error. pl edit it..
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A coin is weighted so that the probability of heads on any flip is 0.6  [#permalink]

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New post 19 Sep 2015, 06:29
1
Bunuel wrote:
A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?

A. (0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2
B. (0.6)^5 + 4(0.6)^4(0.4) + 6(0.6)^3(0.4)^2
C. (0.6)^5 + 3(0.6)^4(0.4) + 2(0.6)^3(0.4)^2
D. (0.6)^5 + 2(0.6)^4(0.4) + (0.6)^3(0.4)^2
E. (0.6)^5 + (0.6)^4(0.4) + (0.6)^3(0.4)^2

Kudos for a correct solution.


the formula for these Qs is..
nCr*(p)^r*(1-p)^(n-r)...
A is the answer
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A coin is weighted so that the probability of heads on any flip is 0.6  [#permalink]

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New post 20 Sep 2015, 20:01
1
1
Bunuel wrote:
A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?

A. (0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2
B. (0.6)^5 + 4(0.6)^4(0.4) + 6(0.6)^3(0.4)^2
C. (0.6)^5 + 3(0.6)^4(0.4) + 2(0.6)^3(0.4)^2
D. (0.6)^5 + 2(0.6)^4(0.4) + (0.6)^3(0.4)^2
E. (0.6)^5 + (0.6)^4(0.4) + (0.6)^3(0.4)^2

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

In this probability problem, we first need to split up the successful cases into three batches:
a) Flipping zero tails
b) Flipping exactly one tails
c) Flipping exactly two tails

a) Flipping zero tails is the same as flipping five heads in a row (HHHHH). Since the probability of flipping one heads is 0.6, and the flips are independent, we get \((0.6)^5\). Unfortunately, this term is common to every answer choice, perhaps not surprisingly.

b) Flipping exactly one tails is the same as flipping four heads and one tails. These can come in any of five orders, depending on the position of the “odd man out”:
THHHH
HTHHH
HHTHH
HHHTH
HHHHT

Each of these sub-cases has a probability of \((0.6)^4*(0.4)\). Since we have five separate sub-cases, any of which counts as success, we add up the probabilities to get \(5*(0.6)^4*(0.4)\). There is only one answer choice that fits: we could stop here.

For the sake of completeness, let’s do the last part.
c) Flipping exactly two tails is the same as flipping three heads and two tails in any order. To count the arrangements, we can either write them all out or use a combinatorics approach, such as the anagram method. Using the latter, we get \(\frac{5!}{(3!2!)}= 10\) arrangements. This 10 becomes the coefficient on the probability of one particular arrangement (such as HHTHT). We get \(10*(0.6)^3*(0.4)^2\) as the last term.

The correct answer is A.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: A coin is weighted so that the probability of heads on any flip is 0.6  [#permalink]

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Re: A coin is weighted so that the probability of heads on any flip is 0.6 &nbs [#permalink] 01 Mar 2018, 09:12
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