A fair coin is tossed 4 times. What is the probability of getting at : Problem Solving (PS)

gmihir

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

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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

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Originally posted by gmihir on 30 Apr 2012, 01:05.
Last edited by Bunuel on 05 Feb 2019, 05:16, edited 1 time in total.

Renamed the topic.

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Bunuel

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

30 Apr 2012, 01:17

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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

Let's find the probability of the opposite event and subtract this value from 1.

The opposite event would be getting zero tails (so all heads) or 1 tail.

\(P(HHHH)=(\frac{1}{2})^4=\frac{1}{16}\).

\(P(THHH)=\frac{4!}{3!}*(\frac{1}{2})^4=\frac{4}{16}\), we are multiplying by \(\frac{4!}{3!}\) since THHH scenario can occur in number of ways: THHH, HTHH , HHTH, or HHHT (notice that \(\frac{4!}{3!}\) basically gives number of arrangements of 4 letters THHH out of which 3 H's are identcal).

\(P(T\geq{2})=1-(\frac{1}{16}+\frac{4}{16})=\frac{11}{16}\).

Answer: D.

Hope it's clear.
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Re: A fair coin is tossed 4 times. What is the probability of getting at [#permalink]

01 Sep 2013, 23:31

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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

Here's another approach that I used:

Total possibilities = 2^4= 16 since each toss leads to 2 outcomes.

Way of getting no tail in 4 tosses = 1 (H H H H)
Way of getting 1 tail in 4 tosses = 4 (T H H H) (H T H H) ( H H T H) (H H H T)
So ways of getting at least two tails in 4 tosses = 11 (16-5)
Therefore, the probability = 11/16 (D).

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

01 Sep 2013, 19:41

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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

mathivanan

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

14 Aug 2015, 00:39

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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]

08 Dec 2016, 00:47

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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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hannahkagalwala

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

16 Apr 2017, 00:40

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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]

22 May 2017, 09:10

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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]

04 Dec 2017, 04:12

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

12 Jan 2018, 07:31

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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.

Answer: D
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Funsho84

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

12 Jan 2018, 15:22

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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
Answer D

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

25 Jun 2019, 12:47

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?

G

CEdward

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

29 Apr 2021, 17:15

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]

05 May 2021, 04:41

Top Contributor

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}

Probability of having all heads or one head

= 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]

04 Oct 2022, 03:39

Expert Reply

Top Contributor

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 - \(\frac{1}{16}\) - \(\frac{4}{16}\) = \(\frac{16-5}{16}\) = \(\frac{11}{16}\)

So, Answer will be D
Hope it helps!

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