Answer:
Let S be the sample space. Given, E = the event that the card drawn is a spade And F = the event that the card drawn is an ace Then, \[E\cap F=\] the event that the card drawn is an ace and spade. Total number of cards = 52, number of spade cards = 13, Number of ace cards = 4 And number of ace of spade card = 1 i.e. \[n(S){{=}^{52}}{{C}_{1}}=52,\] \[n(E){{=}^{13}}{{C}_{1}}=13,\] \[n(F){{=}^{4}}{{C}_{1}}=4,\] \[n(E\,\cap F)=1\] Now, \[P(E\,\cap F)=\frac{n(E\,\cap F)}{n(S)}=\frac{1}{52}\] \[P(E)=\frac{n(E)}{n(S)}=\frac{13}{52}=\frac{1}{4}\] and \[P(F)=\frac{n(F)}{n(S)}=\frac{4}{52}=\frac{1}{13}\] Here, we see that \[P(E\,\cap F)=\frac{1}{52}=P(E).P(F)\] Hence, the events E and F are independent.
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