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Total cards = 8 + 5 = 13
The selections are being made with replacement. Hence,
a) P(G1 and G2) = P(G1)*P(G2) = (8/13)*(8/13) = 0.3787
b) P(Atleast 1 green) = 1 - P(No green) = 1 - (5/13)*(5/13) = 0.8521
c) P(G2 | G1) = P(G2) = 8/13 = 0.6154
Probability with replacement is used for questions where the outcomes are back again to the sample space again. which means that when the object is selected, then its miles changed and returned to the sample space, so the range of elements of the sample space remains unchanged.
Conclusion: The chance of two consecutive attracts without a replacement from a deck of playing cards is calculated because the range of viable successes over the variety of viable results, is expanded together for each case. accordingly, for the primary ace, there may be a 4/52 possibility and for the second there is a 3/51 probability.
Suppose that you have 8 green cards and 5 yellow cards. The cards are well shuffled. You randomly draw two cards with replacements. Round your answers to four decimal places.
G1 = the first card drawn is green
G2 = the second card drawn is green
a. P(G1 and G2) =
b. P(At least 1 green) =
c. P(G2|G1) =
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