exercise:6fbfccbd30: Difference between revisions
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(Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Two cards are drawn successively from a deck of 52 cards. Find the probability that the second card is higher in rank than the first card. '' Hint'': Show that <math>1 = P(\mbox{higher}) + P(\mbox{lower}) + P(\mbox{same})</math> and use the fact...") |
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Two cards are drawn successively from a deck of 52 cards. Find the probability that the second card is higher in rank than the first card. '' Hint'': Show that <math>1 = P(\mbox{higher}) + P(\mbox{lower}) + P(\mbox{same})</math> and use the fact that <math>P(\mbox{higher}) = P(\mbox{lower})</math>. | |||
cards. Find the | |||
probability that the second card is higher in rank than the first card. '' Hint'': Show that <math>1 = P(\mbox{higher}) + P(\mbox{lower}) + P(\mbox{same})</math> | |||
and use the fact that <math>P(\mbox{higher}) = P(\mbox{lower})</math>. | |||
Latest revision as of 21:12, 12 June 2024
Two cards are drawn successively from a deck of 52 cards. Find the probability that the second card is higher in rank than the first card. Hint: Show that [math]1 = P(\mbox{higher}) + P(\mbox{lower}) + P(\mbox{same})[/math] and use the fact that [math]P(\mbox{higher}) = P(\mbox{lower})[/math].