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A coin tossed three times in succession. if \[E\] is the event that there are at least two heads and\[\;F\] is the event in which first throw is a head, then \[P\left( {\dfrac{E}{F}} \right)\]is equal to :

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Hint:First we toss a coin, it will give 2 outcomes. They are Head or tail,
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 \[P(F) \ne 0\]

Complete step-by-step answer:
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 \[\dfrac{3}{4}\].

Note:If we toss a coin once the probability of getting a head or tail will be the same \[ = \dfrac{1}{2}\].When we toss a coin n times, we will get the total number of outcomes \[ = {2^n}\]
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, \[P\left( {\dfrac{E}{F}} \right) = \dfrac{{P(E \cap F)}}{{P(F)}}\].