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Here we tossed a coin 3 times so it will give \[8\] outcomes. Now we find the possibilities of getting Event \[E\] and Event \[F\] and finally we find probability of \[E\] and \[F\].Here we use to find conditional probability, that is \[P\left( {\dfrac{E}{F}} \right) = \dfrac{{P(E \cap F)}}{{P(F)}}\] , if \[P(F) \ne 0\]

It is given that A coin is tossed three times in succession.

\[E\]: event that there are at least \[2\] heads.

\[F\]: event in which the first throw is head.

Total outcomes

1) H H H

2) H H T

3) H T H

4) T H H

5) T T H

6) T H T

7) H T T

8) T T T

After tossing a coin three times we get \[8\] outcomes

\[F\]: event in which the first throw is head

Total possibilities are \[4\]

H H T or H T H or H T T or H H H

The probability of the event F of getting head in the first trial is independent of the other \[2\] trails so it becomes

\[P(F) = \dfrac{4}{8} = \dfrac{1}{2}\]

\[E\]: event that there are at least \[2\] heads

\[2\] heads and \[1\] tail or \[3\] heads

Now we find \[P(E \cap F)\],

The probability that the \[3\] trails have at least \[2\] heads when the first thrown is a head we can find it

Total possibilities are \[3\]

H H T or H T H or H H H

\[P(E \cap F) = \dfrac{3}{8}\]

\[P(E/F) = \dfrac{{P(E \cap F)}}{{P(F)}}\]

\[P(E/F) = \dfrac{{3/8}}{{1/2}} = \dfrac{3}{4}\]

Therefore, \[P\left( {E/F} \right)\] is equal to \[\dfrac{3}{4}\].

Conditional Probability: If \[E\] and \[F\] are any two event in a sample space \[S\] and \[P(F) \ne 0\], then the conditional probability of given is, \[P\left( {\dfrac{E}{F}} \right) = \dfrac{{P(E \cap F)}}{{P(F)}}\].