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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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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

ShowHide Answer

Official Answer

C,D,E,F

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 13156 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 13156 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.

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

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

Indicate all possible values.

[A] 0.1

[B] 0.2

[C] 0.3

[D] 0.4

[E] 0.6

[F] 0.7

Show: ::

LATER

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Joined: **10 Apr 2015 **

Posts: **6218**

Given Kudos: **136 **

Re: A weighted coin has a probability p of showing heads. If suc
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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
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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

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Posts: **1101**

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GPA: **2.8**

WE:**Engineering (Computer Software)**

Re: A weighted coin has a probability p of showing heads. If suc
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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

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

A weighted coin has a probability p of showing heads. If suc
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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.

Re: A weighted coin has a probability p of showing heads. If suc
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27 Dec 2023, 12:57

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Re: A weighted coin has a probability p of showing heads. If suc [#permalink]

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