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# A fair coin is tossed 4 times. What is the probability of getting at

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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]
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gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

Another method:

Let's find the probability of 2 tails , 3 tails and all 4 tails

P(TTTT)=((1/2)^4=1/16.

P(HTTT)=(4!/3!)*(1/2)^4=4/16, we are multiplying by 4C3 (notice that {4!/3!} basically gives number of arrangements of 4 letters HTTT out of which 3 T's are identcal).

P(TTHH) = 4C2*(1/2)^4 = 4!/2!*2!*(1/2)^4=6/16

Total Probablity = 1/16 + 4/16 + 6/16
=11/16
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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]
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gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

Total outcome is 2^4=16
Number of cases of( tails < or =2) = 5, HHHH, THHH, HTHH, HHTH, HHHT
Therefore, the required probability is 11/16
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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]
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Probability of 2T and 2H = 4C2*1/16 = 6/16
Probability of 3T and 1H = 4C3*1/16 = 4/16
Probability of 4T and 0H = 4C4*1/16 = 1/16

Summing the probabilities for each scenario that fulfils our requirements = 11/16
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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]
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gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

This is how I solved it, please tell me if it is correct.

Let's list the possibilities:
HHTT
HTHT
THHT
TTHH
THTH
HTTH

TTTH
TTHT
THTT
HTTT

TTTT

Now let's look at the first one: HHTT. Probability of getting H=1/2, H=1/2, T=1/2, T=1/2. So probability of getting HHTT = 1/2*1/2*1/2*1/2 = 1/16.
Similarly, for all the rest, the probability is 1/16. There are 11 possibilities on the list so the answer is 11*1/16 = 11/16.
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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]
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This is a classic. it should be approach with a general concept.
Otherwise, it is a waste of time to reconstruct the logic everytime we face the same type of question.

The concept is binomial probabilities.

Proba of having k success out-of n events for a binomial variable has a given formula.
Let say, proba of success = p and proba of failure = (1-p)

Formula is nCk * (p^k)*(1-p)^(n-k)
if p=1/2 then formula becomes nCk/(2^n)

In the current question, p=1/2 and (1-p)=1/2
n=4 and p=2 or 3 or 4

Answer is (4C2+ 4C3 + 4C4)/(2^4) = 11/16

Option - D
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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]
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gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

I eliminated the options like follows:

A coin is tossed 4 times,
Probability of getting "at least 2" can be getting 2 tails, 3 tails and 4 tails.

Thus, probability >0.5.

There is only one option with probability >0.5

Please correct me if I am wrong, and yes I agree this is very crude approach.

I am very weak in probability.
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gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

P(at least 2 tails) = 1 - P(not getting at least 2 tails)

P(at least 2 tails) = 1 - [P(getting exactly 1 tail) + P(getting 0 tails)]

Let’s look at P(getting exactly 1 tail):

P(T - H - H - H) = (1/2)^4 = 1/16

Since T - H - H - H can be arranged in 4!/3! = 4 ways, then P(getting exactly 1 tail) = 4 x 1/16 = 4/16.

Now let’s look at P(getting 0 tails), i.e., P(getting all 4 heads):

P(H - H - H - H) = (1/2)^4 = 1/16

P(at least 2 tails) = 1 - (4/16 + 1/16) = 11/16.

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Total outcome = 2^4 = 16

3 cases for favorable outcomes: (HHTT), (HTTT), (TTTT)
4C2 + 4C3 + 4C4 => 6 + 4 + 1= 11

Probability of at least 2 = (favorable outcome/ total outcome) = 11/16
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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]
hannahkagalwala wrote:
gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

This is how I solved it, please tell me if it is correct.

Let's list the possibilities:
HHTT
HTHT
THHT
TTHH
THTH
HTTH

TTTH
TTHT
THTT
HTTT

TTTT

Now let's look at the first one: HHTT. Probability of getting H=1/2, H=1/2, T=1/2, T=1/2. So probability of getting HHTT = 1/2*1/2*1/2*1/2 = 1/16.
Similarly, for all the rest, the probability is 1/16. There are 11 possibilities on the list so the answer is 11*1/16 = 11/16.

I don't know how you arrange all these probabilities without making an error. Could you enlighten me?
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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

Possibilities:
P(TTHH) = (1/2)^2 x (1/2)^2 = 1/16...but there is a permutation here... so 1/16 x 4! / 2! x 2! = 6/16
P(TTTH) = (1/2)^3 x (1/2) = 1/16 ....1/16 x 4!/3! = 4/16
P(TTTT) = (1/2)^4 = 1/16

6/16 + 4/16 + 1/16 = 11/16

D.
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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]
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Solution:

At least two tails imply 2 tails,3 tails or all of them tails.

Lets calculate the probability of NOT getting at least 2 tails.

Then,

(i)Probability of getting all heads would be (1/2)^4 =1/16 and

(ii)Probability of getting one Tail and 3 heads would be P(T,H,H,H) = 4!/3! * (1/2)^4

{as THHH can occur as THHH, HTHH , HHTH, or HHHT}

= 1/16 + 4/16

= 5/16

Thus Probability to get at least two tails

= 1- 5/16

=11/16 (option d)

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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]
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Given that A fair coin is tossed 4 times, and we need to find What is the probability of getting at least 2 tails

Coin is tossed 4 times => Total number of cases = $$2^4$$ = 16

At least 2 heads means that we can get 2 or more tails. This is also equal to 1 - P(0T) - P(1T) (As P(0T) + P(1T) + P(2T) + P(3T) + P(4T) = 1)

P(0T) = $$\frac{1}{16}$$, As there is only one case of HHHH when we will get 0T

P(1T) = $$\frac{4}{16}$$, As there are 4 such cases where the Tail can come in any spot. THHH, HTHH, HHTH, HHHT

=> P(At least 2 tail) = 1 - P(0T) - P(1T) = 1 - $$\frac{1}{16}$$ - $$\frac{4}{16}$$ = $$\frac{16-5}{16}$$ = $$\frac{11}{16}$$

Hope it helps!

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

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