Question

Suppose you are asked to draw ten cards without replacement from a regular deck of 52 playing cards. What is the probability of getting exactly 3 Queens or exactly 3 Kings (or both)?

С https://drive.google.com/drive/folders/lyinXTMXuBMbKU3nO0oRbVKV ug pg 1 Yrmu (18) Suppose you are asked to draw 5 cards from a deck of 52 regular playing cards (a) If asked to draw w/ replacement, what is the probability that no two cards will have the same face value? (b) If asked to draw w/o replacement, what is the probability at least 4 of the drawn cards will be of the same suit?6 (19) Suppose from a collection of n objects some r of them are drawn. Sup- pose further that we preselect N objects of this set of n objects. What is the probability that none of the preselected N is included in the drawn r objects if the sampling is (a) without replacement. (b) with replacement (20) Suppose you are asked to draw ten cards without replacement from a regular deck of 52 playing cards. What is the probability of getting Ans: 0.92 (52) (48)(44) (40) (36)/(52) 7:55 PM Type here to search '^ ./аф 3/20/2019

I also need help with the other two, please provide an explanation with your work and I will promptly give a positive rating.

Answer 1

As per HomeworkLib policy only one question can be answered

so i am solving 18th question

Required probability that no two cards have same face value

$=\frac{\binom{13}{1}\binom{4}{1}\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{4}{1}\binom{10}{1}\binom{4}{1}\binom{9}{1}\binom{4}{1}}{52^5}=0.4159$

b ) Required probability at least 4 of cars drawn will be of same suit

$=\frac{\binom{4}{2}\binom{2}{1}\binom{13}{4}\binom{13}{1}}{\binom{52}{5}}+\frac{\binom{4}{1}\binom{13}{5}}{\binom{52}{5}}=0.0449$