11 (6 points). A 5-card hand is dealt from a well-shuffled deck of 52 playing cards....

ASGಶ: Given that deck of 52 A 5-caod hand is dealt from a well-shuffled playing cards selecting 5 cards out of 52 coxds in ( 3 ) ways Let A be the event that there will be at least one card from each of the four suits in a hand. The probability of the above event is calculated from it's complementary event À there will be no cards from at least one of the four suits Let 's be the event of having no spade cards in the section of 5 cards, Hrepresents the event of having no heart cards in the section of 5 cards. D' represents the event of having no diamond cards in the section of 5 cards, and c represents the event of having no club cards in the section of 5 cards. Find the probability of having no cards from atleast one of the four soits in a hand that means, find the value of PCA) = (SUHUDUC). By the additional theorem of probability, we can write the required probability as follows.

P(SUHU DUC) = P(S) +PCH) +PCD) +PC) -PCSH)- P(SD)- P(SC) - PCHD)-PCHC) - PCDC) +PGHD) +PCSHC)+ PCHDC) - P(Sn Hoone -(29)-(5)-(5)-(3) - (36)-(65)+)-(3) 4(32) - 6( 26 ) + 4( 3 (4x5,75, 757) - [6 x65,786) +(4x1,287) 2,598, 960 . 1913.9 6.co 1,913.296 2,598.960 = 0.7363 (PCA) 0.7363

Thus, the probability having no cards from atleast one of the four suits in a hand is PLAC) = 0.7363 Therefoto Therefore, the probability that there will be at least one caxd from each of the four suits in a hand is given as PCA) - 1-PCA) = 1-0.7363 = 0.2637 F-PCA) : 0.2637 - Thank you -