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A coin is tossed seven times.

Let,

Event A is the head obtained on the 1^st toss.

Event B is the head obtained on the 5^th toss.

For Event A and B to be disjoint P (A) should not equal to P(B) .

To get the probability of event A and B

A coin is tossed 7 times. Thus,\(n\left( s \right) = 128\)

To get the probability of the event that a head is obtained on 1^st toss, we have to consider the variable x that denotes heads appears in 1^st position. Here we have total 1, 2....,7 positions. Each position is a success or failure

Hence, it models binomial distribution with parameter 7 and \({\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.}\!\lower0.7ex\hbox{$2$}}\)

Therefore

\({\rm P}\left( {\rm A} \right) = 7\)

Similarly calculate the probability of event B

\({\rm P}\left( {\rm B} \right) = 7\)

Using the probabilities computed above and substituting in the required condition, we get

It shows,

\(\)\({\rm P}\left( {\rm A} \right) = {\rm P}\left( {\rm B} \right)\)

Since every outcome is independent event A and B are not disjoint.