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

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

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

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

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