I toss three coins together. The possible out | vedclass

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$(D)$ When tossing three coins together,the total number of possible outcomes is $2^3 = 8$.These outcomes are: $(HHH), (HHT), (HTH), (THH), (HTT), (TH

Vedclass · Accepted Answer

Both $A$ and $C$.

Vedclass · Answer

$(D)$ When tossing three coins together,the total number of possible outcomes is $2^3 = 8$.These outcomes are: $(HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT)$.Each of these $8$ outcomes is equally likely.The event of getting 'no heads' corresponds to only one outcome: $(TTT)$.Therefore,the probability of getting no heads is $\frac{1}{8}$.The conclusion that the probability is $\frac{1}{4}$ is incorrect because it assumes that the four outcomes (no heads,$1$ head,$2$ heads,$3$ heads) are equally likely,which is not true.

$I$ toss three coins together. The possible outcomes are no heads,$1$ head,$2$ heads and $3$ heads. So,$I$ say that probability of no heads is $\frac{1}{4}$. What is wrong with this conclusion?

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