If the coin has equal probability of landing heads up or tai

[Guest]

If the coin has equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exactly twice in 3 consecutive flips?
1) 0.125
2) 0.25
3) 0.375
4) 0.5
5) 0.666

Ans- 0.375

Source : Gmatprep 2

[Saurav]

Case 1

H-H-T = 1/2 * 1/2 * 1/2

Case 2
H-T-H = 1/2 * 1/2 * 1/2

Case 3
T-H-H = 1/2 * 1/2 * 1/2

Total = Case 1 + 2 + 3 = 3 * 1/8 = 3 * 0.125

[RonPurewal]

Case 1

H-H-T = 1/2 * 1/2 * 1/2

Case 2
H-T-H = 1/2 * 1/2 * 1/2

Case 3
T-H-H = 1/2 * 1/2 * 1/2

Total = Case 1 + 2 + 3 = 3 * 1/8 = 3 * 0.125

correct.

here are two other ways you can approach this problem.

(1) basic formula for probability: (number of successful outcomes) / (TOTAL number of outcomes), provided that the "outcomes" are equally likely:
just list all the outcomes. there are only eight of them, so, if you're organized, you shouldn't have any trouble creating the list in a moderate amount of time.
hhh
hht
hth
thh
htt
tht
tth
ttt
of the 8 possibilities in the list, each of which is equally likely, there are 3 (hht, hth, thh) that satisy the criterion in the problem. therefore, the desired probability is 3/8.

--

(2) if you really like formulas, you can use the formula for probability in Bernoulli trials, as detailed in this thread.
that gives (3!/2!1!) x (1/2)^2 x (1/2)^1, which reduces to 3/8.

[bamalwa]

Hi Ron,

Can we use this approach---

there are two possibilities for first coin flip - (H and T)
Same goes for second and third flips

So total possibilities = 2*2*2 = 8

there are three ways we can get two heads out of three flips -
HHT, HTH, THH.

Probability = 3/8 = 0.125*3 = 0.375

[cseramit]

Hi Ron,

Can we use this approach---

there are two possibilities for first coin flip - (H and T)
Same goes for second and third flips

So total possibilities = 2*2*2 = 8

there are three ways we can get two heads out of three flips -
HHT, HTH, THH.

Probability = 3/8 = 0.125*3 = 0.375

@bamalwa,
Your method is nothing but the "basic formula for probability" that Ron has mentioned in his previous post.

Thanks
Amit

[RonPurewal]

@bamalwa,
Your method is nothing but the "basic formula for probability" that Ron has mentioned in his previous post.

Thanks
Amit

true, but there is one difference -- this poster has calculated the total number of outcomes for the situation, while i found that number by just listing all the outcomes. both are valid.

[tsiria]

Can you use a combination of combinatorics/permutations?

There are 2 choices for each flip: so total of 3 flips = 2*2*2 = 8 different configurations.

Different arrangement of two heads: 3C2 : 3!/2! = 3

Therefore probability is: 3/8 = 0.375

[RonPurewal]

Can you use a combination of combinatorics/permutations?

There are 2 choices for each flip: so total of 3 flips = 2*2*2 = 8 different configurations.

Different arrangement of two heads: 3C2 : 3!/2! = 3

Therefore probability is: 3/8 = 0.375

yep.

[amit1234]

hi Ron,

how to solve this by slot method?

here repetition is not allowed so

favorable outcome =
3*2*1/2*1 =3

how to get total outcome....?

[RonPurewal]

hi Ron,

how to solve this by slot method?

well, "slot method" is for combinatorics, not probability.

if you mean "how do i solve this by multiplying probabilities", then you have to list out the three different orders in which the desired event could happen (HHT, HTH, THH).
each of those is 1/2 x 1/2 x 1/2 = 1/8, so the total probability is 1/8 + 1/8 + 1/8 = 3/8.