State true or false:Three different coins are tossed together. The probability of getting at least two heads is 12.

# State true or false:Three different coins are tossed together. The probability of getting at least two heads is $\frac{1}{2}$.

1. A
True
2. B
False

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### Solution:

It is given that three different coins are tossed together.
We know that the probability is given as the ratio of the number of favorable outcomes with the total number of possible outcomes.
$P\left(E\right)=\frac{\mathit{Number of favourable outcomes n}\left(E\right)}{\mathit{Total possible outcomes n}\left(S\right)}$
When we flip three coins simultaneously the chance of results are HHH or HHT or HTH or THH or HTT or THT or TTH or TTT individually; where H is meant for the head and T is signified for the tail.
So the sample space for three coins tossed is:
$S=\left\{\mathit{HHH},\mathit{HHT},\mathit{HTH},\mathit{THH},\mathit{HTT},\mathit{THT},\mathit{TTH},\mathit{TTT}\right\}$.
$⇒n\left(S\right)=8$
Let E be the event of getting atleast two heads.
$E=\left\{\mathit{HHH},\mathit{HHT},\mathit{HTH},\mathit{THH}\right\}$
$⇒n\left(E\right)=4$
The probability of event E is,
$P\left(E\right)=\frac{\mathit{Number of favourable outcomes n}\left(E\right)}{\mathit{Total number of outcomes n}\left(S\right)}$
$⇒P\left(E\right)=\frac{4}{8}$
$⇒P\left(E\right)=\frac{1}{2}$
Thus, the probability of getting at least two heads $\frac{1}{2}$.
Therefore, the given statement is true.
Hence, option 1 is correct.

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