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Home » Two coins are tossed once, where (i) E: tail appears on one coin, : one coin shows head (ii) E: no tail appears, F: no head appears
Two coins are tossed once, where (i) E: tail appears on one coin, \mathbf{F} : one coin shows head
(ii) E: no tail appears, F: no head appears
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Two coins are tossed once, where (i) E: tail appears on one coin, \mathbf{F} : one coin shows head
(ii) E: no tail appears, F: no head appears

Solution:

Determining the sample space of the given experiment is \mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \Pi\}\}

(i) Given that, E: tail appears on one coin

And F: one coin shows head

Sample space of events,

\Rightarrow \mathrm{E}=\{\mathrm{HT}, \mathrm{TH}\} and \mathrm{F}=\{\mathrm{HT}, \mathrm{TH}\}

\text{E}\cap \text{F}=\{\text{HT},\text{TH}\}

The value of probability of events,  P(E)=\frac{2}{4}=\frac{1}{2}, P(F)=\frac{2}{4}=\frac{1}{2}, P(E \cap F)=\frac{2}{4}=\frac{1}{2}

Now, we know that conditional probability is given by relation:

P(E \mid F)=\frac{P(E \cap F)}{P(F)}

Substituting the values, we get

\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{1 / 2}{1 / 2}

\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=1

(ii) Given, E: no tail appears

And F: no head appears

Sample space of events,

\Rightarrow \mathrm{E}=\{\mathrm{HH}\} and \mathrm{F}=\{T \mathrm{~T}\}

So common sample space,

\Rightarrow E \cap F=\varphi

 The value of probability of events,

 P(E)=\frac{1}{4}, P(F)=\frac{1}{4}, P(E \cap F)=\frac{0}{4}=0

Now, we know that conditional probability relation is,

P(E \mid F)=\frac{P(E \cap F)}{P(F)}

Substituting the values, we get

    \[\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{0}{1 / 4}\]

\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=0

i

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