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P(A and B get same number of heads) = P(A and  B both get 0 head ) + P(A and B both get 1 head) +  P(A and B both get 1 head) +  P(A and B both get 1 head)                                      .................(1)

Now P(A and  B both get 0 head ) :   3C0 (1/2)3 . 3C0 (1/2)3 [Since A and B toss 3 coins each separately so they need to be considered separate events.Also since it is a coin toss problem so it considers the requirements of binomial distribution and hence we apply this formula but separately for A and B and multiplying them since both of them will occur hence follows multiplication principle and also A and B are separate events as mentioned earlier]

                                                 =  1/64

SImilarly P(A and  B both get 1 head ) :  3C1 (1/2)3 . 3C1 (1/2)3

                                                                  =  9/64

SImilarly P(A and  B both get 2 heads ) :  3C2 (1/2)3 . 3C2 (1/2)3

                                                                  =  9/64

SImilarly P(A and  B both get 3 heads) :  3C3 (1/2)3 . 3C3 (1/2)3

                                                                  =  1/64

Hence from (1) , we have :

P(A and B get same number of heads)  =  1/64 + 9/64 + 9/64 + 1/64

                                                             = 20/64

                                                             =  5/16

Hence , C) should be the correct option.

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