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

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

As an independent event & 2 successive flips.

Checking Given Option:
1. Probability = 0.3 , Prob. of getting 1 head = 1 - (1-0.3)^2 = 0.51
2. Probability = 0.4 , Prob. of getting 1 head = 1 - (1-0.4)^2 = 0.64
3. Probability = 0.7 , Prob. of getting 1 head = 1 - (1-0.7)^2 = 0.91

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

I. 0.3
II. 0.4
III. 0.7

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

Here
p(1 - p) + p * p > 0.5
p - p^2 + p^2 > 0.5
p > 0.5
0.7 > 0.5
Only III satisfies the condition.

ALTERNATIVELY,

p(1 - p) + p * p > 0.5
I. p = 0.3; 0.3*0.7 + 0.3*0.3 = 0.3
II. p = 0.4; 0.4*0.6 + 0.4*0.4 = 0.4
III. p = 0.7; 0.7*0.3 + 0.7*0.7 = 0.7

Only III satisfies.

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

Given

• The probability of showing heads for the weighted coin is ‘p’
o Please note, weighted coin means the probability of getting head and tail can be different, the term ‘biased’ is also used for such coins.
• Successive flips are independent.

To Find
• Which of the options given could be a value of ‘p’. Give, the probability of getting at least one head in two flips is greater than 0.5

Approach and Working Out

Probability of getting a head = p
• Hence, Probability of getting a tail = $$(1 – p)$$

In two flips probability of getting at least one head
= 1 - Probability of getting no heads
= 1 - the probability of getting two tails.

Probability of getting two tails = $$(1 – p)^2$$

Hence, 1 – $$(1-p)^2$$ > 0.5

Now, we can check by the options given,
I. 1 – $$(1 – 0.3)^2$$ = 1 – $$0.7^2$$ = 0.51 (Valid)
II. 1 – $$(1 – 0.4)^2$$ = 1 – $$0.6^2$$ = 0.64 (Valid)
III. 1 – $$(1 – 0.5)^2$$ = 1 – $$0.5^2$$ = 0.75 (Valid)

All three values are possible.

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Re: A weighted coin has a probability p of showing heads. If successive fl [#permalink]
IMO C.

The question indirectly says that the coin is biased.

The probability of getting one head in two successive flips is greater than 0.5
i.e.

HH
HT
TH

probability of getting TT is less than 0.5

Therefore the probability of getting atleast one head should be atleast greater than 0.5

The only option is C (0.7)

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Re: A weighted coin has a probability p of showing heads. If successive fl [#permalink]
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=> $$P(TT) < 0.5$$
=> $$T^2 < 0.5$$
=> $$(1-H)^2 < 0.5$$
=> $$1-H < \sqrt{0.5}$$
=> $$H > 1 - \sqrt{0.5}$$

$$\sqrt{0.5} > 0.7$$

=> $$1 - \sqrt{0.5} < 0.3$$
=> H can take all possible values of 0.3, 0.4, and 0.7

Ans: E
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Re: A weighted coin has a probability p of showing heads. If successive fl [#permalink]
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If p = 3/10, the probability of getting Tails on one flip is 7/10, and the probability of getting Tails twice in a row on two flips would be (7/10)(7/10) = 49/100, so the probability of getting Heads at least once would be 0.51. So p can be 0.3. Clearly p can then also be anything greater than 0.3, so the answer is E.
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Re: A weighted coin has a probability p of showing heads. If successive fl [#permalink]
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All tails = $$(1-p)*(1-p) = (1-p)^2$$since events are independent
Atleast 1 head = $$1-(1-p)^2$$
Now, $$1-(1-p)^2 > 0.5$$

When p=0.3, LHS=0.51>0.50
Case is valid

When p=0.4, LHS=0.64>0.50
Case is valid

When p=0.7, LHS=0.91>0.50
Case is valid

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

the probability of getting heads on at least 2 flips is less than 0.5 on either of the answers if they are dependent.

0.7^2 = 0.49
0.4^2 = 0.16
0.3^2 = 0.09

But since they're independent, then only 0.7 has a probability of getting heads that is over 0.5

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Re: A weighted coin has a probability p of showing heads. If successive fl [#permalink]
When we flip 2 baised coins , the possible combinations are {HH, HT,TH,TT} and in the question above we are asked the to find the probability of getting atleast one head so we will have an equation
p(1-p)+(p-1)p+p*p >0.5.
I.e. probability of [HT,TH and HH]

Solving this quadratic equality , we get the possible values of p as
1+-1/√2. Since "p" is the probability, hence we take the answer less than 1 and therefore the value of p will be
0.293<p<1.
Hence all the above could be the answer.

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Re: A weighted coin has a probability p of showing heads. If successive fl [#permalink]
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IMO E

P(H) = p
Total Flips = 2
P(at least one Head) = 1- P(No head) = 1- P(TT)= 1- (1-p)^2

A/Q, 1-(1-p)^2 > 0.5
(1-p)^2 < 0.5
- √0.5 < 1-p < √0.5
-0.707 < 1-p < 0.707
-1.707 < -p < -0.29
0.29 < p < 1.707
But p can be max = 1,
So, 0.29 < p < 1

I. 0.3
II. 0.4
III. 0.7
All three satisfy .

E. I, II, and III
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Re: A weighted coin has a probability p of showing heads. If successive fl [#permalink]
Ans- C

Probability of atleast one heads in two flips= One p + both p = p(1-p) + p x p > .5

This give p> .5

So only III is correct.

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Re: A weighted coin has a probability p of showing heads. If successive fl [#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