Also what was given in the official answer key? Can anyone tell me.

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Suppose three coins are lying on a table, two of them with heads facing up and one with tails facing up. One coin is chosen at random and flipped. What is the probability that after the flip the majority of the coins(i.e., at least two of them) will have heads facing up?

- $\left(\frac{1}{3}\right)$
- $\left(\frac{1}{8}\right)$
- $\left(\frac{1}{4}\right)$
- $\left(\frac{1}{4}+\frac{1}{8}\right)$
- $\left(\frac{2}{3}\right)$

According to me "flip" means that if the chosen coin was head, it will be placed as tail and vice versa. But in the given solution, it is being considered as "tossed".. If "flip" is being considered as "toss", then what is being considered as "flip"? I am confused.

Also what was given in the official answer key? Can anyone tell me.

Also what was given in the official answer key? Can anyone tell me.

0

26 votes

Best answer

**(e) is correct**

Table has $3$ coins with $H,H,T$ facing up.

Now, probability of choosing any coin is $\dfrac{1}{3}$ , as we can chose any of the three coins.

**Case A**: $1^{st}$ coin : either $H$ or $T$ can come.

so, $HHT ,THT$ possible.only $HHT$ is favorable.

which gives $\left(\dfrac{1}{3}\right)\times \left(\dfrac{1}{2}\right)= \dfrac{1}{6}.$

**Case B:** $2nd$ coin : either $H$ or $T$ can come.

so, $HHT, HTT$ possible.only $HHT$ is favourable.

which gives $\left(\dfrac{1}{3}\right)\times \left(\dfrac{1}{2}\right)=\dfrac {1}{6}.$

**Case C:** $3rd$ coin : Table already contains two $H's$ so, whatever comes is favourable.

which gives $\left(\dfrac{1}{3}\right)\times 1 =\dfrac {1}{3}.$

Summing up the total gives $\dfrac{1}{6} +\dfrac {1}{6} +\dfrac {1}{3} =\dfrac {2}{3}.$

__Case1) Lucky but not so lucky (we got the coin facing 'H' up, and the other 2 coins are H and T) __

Probability of getting 'H' $\cap$ Probability of tossing the coin got and getting the result as head = (2/3) * (1/2) = (1/3)

**Case2) Very lucky :P (2 heads are on the table and luckily we got the coin facing 'T' up) **

Probability of getting 'T' $\cap$ Probability of tossing the coin got and getting anything as it does not matter (because we already have 2 other coins which are both heads) = (1/3) * (2/2) = (1/3)

Summing the probability of the above 2 cases = (1/3) + (1/3) = (2/3)

0

6 votes

We have coin lying on table.

we have to select one coin ad then toss it.

In starting **we have {H,H,T } outcome**

Now from here we can only select one coin then flip it.

So,number of all posible outcomes are those which have** 1 or 0 hamming distance from above outcome.**

Here total outcomes are ---6

And outcomes with at least two heads are----4

**Ans is=4/6=2/3**

1 vote

For selecting a head faces coin probability is $2/3$

After selecting one head facing coin the probability that it changes to a tail after the toss is $1/2$

So the probability for not getting majority of the coins not facing head is $((2/3)*(1/2))=1/3$

From this the majority of the coins facing head is $ =1-(1/3)$

$ =2/3$

After selecting one head facing coin the probability that it changes to a tail after the toss is $1/2$

So the probability for not getting majority of the coins not facing head is $((2/3)*(1/2))=1/3$

From this the majority of the coins facing head is $ =1-(1/3)$

$ =2/3$