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A weighted coin has a probability p of showing heads. If suc
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01 Dec 2018, 10:54

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Question Stats:

A weighted coin has a probability p of showing heads. If successive flips are independent, and the probability of getting at least one head in two flips is greater than 0.5, then what could p be?

Indicate all possible values.

A. 0.1

B. 0.2

C. 0.3

D. 0.4

E. 0.6

F. 0.7

_________________

Indicate all possible values.

A. 0.1

B. 0.2

C. 0.3

D. 0.4

E. 0.6

F. 0.7

_________________

Re: A weighted coin has a probability p of showing heads. If suc
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15 Feb 2020, 11:19

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Official Explanation:

Remember that, to calculate the probability of an "at least" scenario, we use the complement rule. The case we want is (at least one H in two flips). The complement of that is (no H in two flips). If p is the probability of H, then (1 – p) is the probability of T. The probability of two T in two flips would be that squared, and then we would subtract from 1 to find the "at least" probability. In the table below, the first column is the possible values of p, the probability of getting H on a single flip. The second column is the probability of getting T on a single flip. The third column is the probability of getting two T's in a row, i.e. no H in two flips; that is the complement of the "at least" case. The final column is the probability of "at least one H in two flips."

Screenshot from 2020-02-16 01-20-29.png [ 86.07 KiB | Viewed 11893 times ]

We see that for all value of p ≥ 0.3, the "at least" probability is greater than 0.5.

FAQ: Why is the (1 - p) term being squared?

We know that:

P(heads) = p

P(tails) = (1 - p)

This problem ultimately asks us to find the probability of getting "at least one heads in two flips". This means that we want to find the probability of getting the following outcomes:

heads, heads

OR

heads, tails

OR

tails, heads

We can calculate all this more easily by first finding the complement to that. The complement to getting "at least one heads in two flips" is getting "exactly 2 tails":

tails, tails

Thus, we're looking for the probability of getting tails AND tails:

P(exactly 2 tails) = P(tails) * P(tails) = P(tails)^2

Substituting in for the value of P(tails), we get:

P(exactly 2 tails) = (1 - p)^2

Taking the complement of this gives us the final expression for our chart:

P(at least one heads) = 1 - P(exactly 2 tails)

P(at least one heads) = 1 - (1 - p)^2

FAQ: Do we really have to make that whole chart? That would take too long!

No, you don't have to fill in that whole chart. That's just being used to illustrate the thinking behind this problem. To solve the problem, you really only need to know that expression in the last column: 1 - (1 - p)^2. We know that that expression must be greater than 0.5. So we end up with:

1 - (1 - p)^2 > 0.5

Now you can just plug in the different answer choices as the value for p in this expression and see which values yield a true statement.

_________________

Remember that, to calculate the probability of an "at least" scenario, we use the complement rule. The case we want is (at least one H in two flips). The complement of that is (no H in two flips). If p is the probability of H, then (1 – p) is the probability of T. The probability of two T in two flips would be that squared, and then we would subtract from 1 to find the "at least" probability. In the table below, the first column is the possible values of p, the probability of getting H on a single flip. The second column is the probability of getting T on a single flip. The third column is the probability of getting two T's in a row, i.e. no H in two flips; that is the complement of the "at least" case. The final column is the probability of "at least one H in two flips."

Attachment:

Screenshot from 2020-02-16 01-20-29.png [ 86.07 KiB | Viewed 11893 times ]

We see that for all value of p ≥ 0.3, the "at least" probability is greater than 0.5.

FAQ: Why is the (1 - p) term being squared?

We know that:

P(heads) = p

P(tails) = (1 - p)

This problem ultimately asks us to find the probability of getting "at least one heads in two flips". This means that we want to find the probability of getting the following outcomes:

heads, heads

OR

heads, tails

OR

tails, heads

We can calculate all this more easily by first finding the complement to that. The complement to getting "at least one heads in two flips" is getting "exactly 2 tails":

tails, tails

Thus, we're looking for the probability of getting tails AND tails:

P(exactly 2 tails) = P(tails) * P(tails) = P(tails)^2

Substituting in for the value of P(tails), we get:

P(exactly 2 tails) = (1 - p)^2

Taking the complement of this gives us the final expression for our chart:

P(at least one heads) = 1 - P(exactly 2 tails)

P(at least one heads) = 1 - (1 - p)^2

FAQ: Do we really have to make that whole chart? That would take too long!

No, you don't have to fill in that whole chart. That's just being used to illustrate the thinking behind this problem. To solve the problem, you really only need to know that expression in the last column: 1 - (1 - p)^2. We know that that expression must be greater than 0.5. So we end up with:

1 - (1 - p)^2 > 0.5

Now you can just plug in the different answer choices as the value for p in this expression and see which values yield a true statement.

_________________

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Re: A weighted coin has a probability p of showing heads. If suc
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04 Dec 2018, 06:34

3

Carcass wrote:

A weighted coin has a probability p of showing heads. If successive flips are independent, and the probability of getting at least one head in two flips is greater than 0.5, then what could p be?

Indicate all possible values.

A. 0.1

B. 0.2

C. 0.3

D. 0.4

E. 0.6

F. 0.7

Indicate all possible values.

A. 0.1

B. 0.2

C. 0.3

D. 0.4

E. 0.6

F. 0.7

Explanation::

Probability of getting one head = 1 - (probability of no head)

Since the events are independent and there are 2 successive flips,

so,

Probability of getting one head = 1 - (probability of no head) (probability of no head) = 1- (probability of no head)^2

Let compare each solution:

A) Probability = 0.1 , probability of getting one head = 1 - (1-0.1)^2 = 0.19 (probability of no head = 1 - probability of getting head)

B) Probability = 0.2 , probability of getting one head = 1 - (1-0.2)^2 = 0.36

C) Probability = 0.3 , probability of getting one head = 1 - (1-0.3)^2 = 0.51

D) Probability = 0.4 , probability of getting one head = 1 - (1-0.4)^2 = 0.64

E) Probability = 0.5 , probability of getting one head = 1 - (1-0.5)^2 = 0.75

F) Probability = 0.6 , probability of getting one head = 1 - (1-0.6)^2 = 0.84

G) Probability = 0.7 , probability of getting one head = 1 - (1-0.7)^2 = 0.91

Hence the options are C,D,E,F,G

_________________

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A weighted coin has a probability p of showing heads. If suc
[#permalink]
17 Feb 2020, 06:53

4

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A weighted coin has a probability p of showing heads. If successive flips are independent, and the probability of getting at least one head in two flips is greater than 0.5, then what could p be?

Indicate all possible values.

[A] 0.1

[B] 0.2

[C] 0.3

[D] 0.4

[E] 0.6

[F] 0.7

_________________

New to the GRE, and GRE CLUB Forum?

GRE: All About GRE | Search GRE Specific Questions | Download Vault

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ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation. | ETS All Official Guides

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Indicate all possible values.

[A] 0.1

[B] 0.2

[C] 0.3

[D] 0.4

[E] 0.6

[F] 0.7

Show: ::

LATER

_________________

GRE: All About GRE | Search GRE Specific Questions | Download Vault

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ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation. | ETS All Official Guides

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Re: A weighted coin has a probability p of showing heads. If suc
[#permalink]
21 Feb 2020, 11:10

huda wrote:

A weighted coin has a probability p of showing heads. If successive flips are independent, and the probability of getting at least one head in two flips is greater than 0.5, then what could p be?

Indicate all possible values.

[A] 0.1

0.2

[C] 0.3

[D] 0.4

[E] 0.6

[F] 0.7

Indicate all possible values.

[A] 0.1

0.2

[C] 0.3

[D] 0.4

[E] 0.6

[F] 0.7

Show: ::

C, D, E, and F

ASIDE: We COULD solve this question by using the complement, but I'm going to keep things straightforward

Given: [b]P(heads) = p

So, P(not heads) = 1 - p

In other words, P(tails) = 1 - p

We want: P(at least 1 heads)

We can rewrite this as: P(at least 1 heads) = P(1st is heads AND 2nd is heads OR 1st is heads AND 2nd is tails OR 1st is tails AND 2nd is heads)

= P(1st is heads AND 2nd is heads) + P(1st is heads AND 2nd is tails) + P(1st is tails AND 2nd is heads)

= [P(1st is heads) x P(2nd is heads)]+ [P(1st is heads) x P(2nd is tails)] + [P(1st is tails) x P(2nd is heads)

= [p x p]+ [p x (1 - p)] + [(1 - p) x p]

= p²+ (p - p²) + (p - p²)

= 2p - p²

GIVEN: P(at least 1 heads) > 0.5

So, we can know write: 2p - p² > 0.5

Or we can rewrite this as: p(2 - p) > 0.5

So we're looking for any values that satisfy the above inequality.

Answer: C, D, E, and F

Cheers,

Brent

_________________

Re: A weighted coin has a probability p of showing heads. If suc
[#permalink]
05 Sep 2022, 06:48

1

p = prob heads, 1-p= prob tails = t

1-t^2 > 0.5 --> -t^2>-0.5 --> t^2<0.5 --> t<0.7071

Since 1-t = p, p>0.2929 --> answer choices C, D, E, F

1-t^2 > 0.5 --> -t^2>-0.5 --> t^2<0.5 --> t<0.7071

Since 1-t = p, p>0.2929 --> answer choices C, D, E, F

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Re: A weighted coin has a probability p of showing heads. If suc
[#permalink]
15 Oct 2022, 09:32

1

Given that A weighted coin has a probability p of showing heads and and the probability of getting at least one head in two flips is greater than 0.5, And we need to find the possible values of p

P(At least one Head) = 1 - P(0 Head) = 1 - P(2 Tail) = 1 - P(TT)

P(H) = p

=> P(T) = 1 - P(H) = 1 - p

=> P(TT) = (1-p) * (1-p) = \((1 - p)^2\)

=> P(At least one Head) = 1 - P(TT) = 1 - \((1 - p)^2\) > 0.5

=> \((1 - p)^2\) < 1 - 0.5

=> \((1 - p)^2\) < 0.5 ~ \((0.7)^2\)

(Watch this video to learn How to Solve Inequality Problems)

=> | 1 - p| ≤ 0.7 (equal to sign because 0.5 is greater than \((0.7)^2\))

(Watch this video to learn the Basics of Absolute Values)

=> -0.7 ≤ 1 - p ≤ 0.7

Multiplying all the sides by -1 (and reverting the sign we get)

=> 0.7 ≥ p - 1 ≥ -0.7

=> -0.7 ≤ p - 1 ≤ 0.7

Adding 1 on all the sides we get

=> -0.7 + 1 ≤ p -1 + 1 ≤ 1 + 0.7

=> 0.3 ≤ p ≤ 1.7

But p cannot be greater than 1

=> All values ≥ 3 are possible

So, Answer will be C, D, E, F

Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

_________________

P(At least one Head) = 1 - P(0 Head) = 1 - P(2 Tail) = 1 - P(TT)

P(H) = p

=> P(T) = 1 - P(H) = 1 - p

=> P(TT) = (1-p) * (1-p) = \((1 - p)^2\)

=> P(At least one Head) = 1 - P(TT) = 1 - \((1 - p)^2\) > 0.5

=> \((1 - p)^2\) < 1 - 0.5

=> \((1 - p)^2\) < 0.5 ~ \((0.7)^2\)

(Watch this video to learn How to Solve Inequality Problems)

=> | 1 - p| ≤ 0.7 (equal to sign because 0.5 is greater than \((0.7)^2\))

(Watch this video to learn the Basics of Absolute Values)

=> -0.7 ≤ 1 - p ≤ 0.7

Multiplying all the sides by -1 (and reverting the sign we get)

=> 0.7 ≥ p - 1 ≥ -0.7

=> -0.7 ≤ p - 1 ≤ 0.7

Adding 1 on all the sides we get

=> -0.7 + 1 ≤ p -1 + 1 ≤ 1 + 0.7

=> 0.3 ≤ p ≤ 1.7

But p cannot be greater than 1

=> All values ≥ 3 are possible

So, Answer will be C, D, E, F

Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

_________________

Ankit

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How to Solve :

Equation of a Line || Co-ordinate Geometry || Remainders || LCM and GCD || Inequalities || Absolute Value || Factors and Multiples || Number and Sum of Factors || Similar Triangles || Exponents || Work Rate || Probability: Coin Toss || Probability: Dice Roll || Prime Numbers || Circles || Equilateral Triangle in a Circle || Arithmetic and Geometric Progression || Divisibility Rules || Statistics || Functions and Custom Characters || Stem & Leaf Plot || Box And Whisker Plot

GRE Club Topic Wise Problems

GRE Quant - 2019 Ed

GRE Verbal - 2019 Ed

The Princeton Review Quantitative Directory

The 5 lb. Book of GRE Practice Problems (2nd Ed.) - QUANT

GRE - PowerPrep® The FREE Practice Tests Explained 2019

Moderator

Joined: **02 Jan 2020 **

Status:**GRE Quant Tutor**

Posts: **1060**

Location: **India**

Concentration: **General Management**

Schools: **XLRI Jamshedpur, India - Class of 2014**

GMAT 1: **700 Q51 V31**

GPA: **2.8**

WE:**Engineering (Computer Software)**

Re: A weighted coin has a probability p of showing heads. If suc
[#permalink]
15 Oct 2022, 09:34

Given that A weighted coin has a probability p of showing heads and and the probability of getting at least one head in two flips is greater than 0.5, And we need to find the possible values of p

P(At least one Head) = 1 - P(0 Head) = 1 - P(2 Tail) = 1 - P(TT)

P(H) = p

=> P(T) = 1 - P(H) = 1 - p

=> P(TT) = (1-p) * (1-p) = \((1 - p)^2\)

=> P(At least one Head) = 1 - P(TT) = 1 - \((1 - p)^2\) > 0.5

=> \((1 - p)^2\) < 1 - 0.5

=> \((1 - p)^2\) < 0.5 ~ \((0.7)^2\)

(Watch this video to learn How to Solve Inequality Problems)

=> | 1 - p| ≤ 0.7 (equal to sign because 0.5 is greater than \((0.7)^2\))

(Watch this video to learn the Basics of Absolute Values)

=> -0.7 ≤ 1 - p ≤ 0.7

Multiplying all the sides by -1 (and reverting the sign we get)

=> 0.7 ≥ p - 1 ≥ -0.7

=> -0.7 ≤ p - 1 ≤ 0.7

Adding 1 on all the sides we get

=> -0.7 + 1 ≤ p -1 + 1 ≤ 1 + 0.7

=> 0.3 ≤ p ≤ 1.7

But p cannot be greater than 1

=> All values ≥ 3 are possible

So, Answer will be C, D, E, F

Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

_________________

Ankit

to BrushMyQuant YouTube Channel to get WEEKLY new VIDEOS!!!

How to Solve :

Equation of a Line || Co-ordinate Geometry || Remainders || LCM and GCD || Inequalities || Absolute Value || Factors and Multiples || Number and Sum of Factors || Similar Triangles || Exponents || Work Rate || Probability: Coin Toss || Probability: Dice Roll || Prime Numbers || Circles || Equilateral Triangle in a Circle || Arithmetic and Geometric Progression || Divisibility Rules || Statistics || Functions and Custom Characters || Stem & Leaf Plot || Box And Whisker Plot

GRE Club Topic Wise Problems

GRE Quant - 2019 Ed

GRE Verbal - 2019 Ed

The Princeton Review Quantitative Directory

The 5 lb. Book of GRE Practice Problems (2nd Ed.) - QUANT

GRE - PowerPrep® The FREE Practice Tests Explained 2019

P(At least one Head) = 1 - P(0 Head) = 1 - P(2 Tail) = 1 - P(TT)

P(H) = p

=> P(T) = 1 - P(H) = 1 - p

=> P(TT) = (1-p) * (1-p) = \((1 - p)^2\)

=> P(At least one Head) = 1 - P(TT) = 1 - \((1 - p)^2\) > 0.5

=> \((1 - p)^2\) < 1 - 0.5

=> \((1 - p)^2\) < 0.5 ~ \((0.7)^2\)

(Watch this video to learn How to Solve Inequality Problems)

=> | 1 - p| ≤ 0.7 (equal to sign because 0.5 is greater than \((0.7)^2\))

(Watch this video to learn the Basics of Absolute Values)

=> -0.7 ≤ 1 - p ≤ 0.7

Multiplying all the sides by -1 (and reverting the sign we get)

=> 0.7 ≥ p - 1 ≥ -0.7

=> -0.7 ≤ p - 1 ≤ 0.7

Adding 1 on all the sides we get

=> -0.7 + 1 ≤ p -1 + 1 ≤ 1 + 0.7

=> 0.3 ≤ p ≤ 1.7

But p cannot be greater than 1

=> All values ≥ 3 are possible

So, Answer will be C, D, E, F

Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

_________________

to BrushMyQuant YouTube Channel to get WEEKLY new VIDEOS!!!

How to Solve :

Equation of a Line || Co-ordinate Geometry || Remainders || LCM and GCD || Inequalities || Absolute Value || Factors and Multiples || Number and Sum of Factors || Similar Triangles || Exponents || Work Rate || Probability: Coin Toss || Probability: Dice Roll || Prime Numbers || Circles || Equilateral Triangle in a Circle || Arithmetic and Geometric Progression || Divisibility Rules || Statistics || Functions and Custom Characters || Stem & Leaf Plot || Box And Whisker Plot

GRE Club Topic Wise Problems

GRE Quant - 2019 Ed

GRE Verbal - 2019 Ed

The Princeton Review Quantitative Directory

The 5 lb. Book of GRE Practice Problems (2nd Ed.) - QUANT

GRE - PowerPrep® The FREE Practice Tests Explained 2019

A weighted coin has a probability p of showing heads. If suc
[#permalink]
26 Nov 2022, 19:59

1

1/2 (chance of getting tails i.e. not heads) * 1/2 = 1/4 chance of not getting heads thus 1- 1/4 = 3/4 chance of getting heads. if the coin has a 50/50 split between heads and tails we know already that E and F are correct. Now we have to plug in numbers for the remaining answer choice

- D: if heads is .4 then tails is .6: 6/10 * 6/10 = 36/10; 1 - 36/100 = 72/100 > 50/100 so D is correct

- C: if heads is .3 then tails is .7: 7/10 * 7/10 = 49/100; 1 - 49/100 = 51/100> 50/100 so C is correct

very unlikely the other choice will be correct given how close C is to being incorrect so CDEF are the answers. Hope the alternative explanation is helpful to someone.

- D: if heads is .4 then tails is .6: 6/10 * 6/10 = 36/10; 1 - 36/100 = 72/100 > 50/100 so D is correct

- C: if heads is .3 then tails is .7: 7/10 * 7/10 = 49/100; 1 - 49/100 = 51/100> 50/100 so C is correct

very unlikely the other choice will be correct given how close C is to being incorrect so CDEF are the answers. Hope the alternative explanation is helpful to someone.

gmatclubot

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