2 votes 2 votes Roll a die, then select at random, without replacement, as many cards from the deck as the number shown on the die. What is the probability that you get at least one Ace? Probability gravner probability engineering-mathematics conditional-probability + – Pooja Khatri asked Sep 23, 2018 Pooja Khatri 668 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes $\frac{1}{6}*[\: (1-\frac{\binom{48}{1}}{\binom{52}{1}}) + (1-\frac{\binom{48}{2}}{\binom{52}{2}}) +(1-\frac{\binom{48}{3}}{\binom{52}{3}}) + (1-\frac{\binom{48}{4}}{\binom{52}{4}}) + (1-\frac{\binom{48}{5}}{\binom{52}{5}}) +(1-\frac{\binom{48}{6}}{\binom{52}{6}})]$ Explanation: P(Ace)=P($1\: on \: dice$)*P($\frac{Ace}{1\: on \: dice}$) + P($2\: on \: dice$)*P($\frac{Ace}{2\: on \: dice}$) + ….. + P($6\: on \: dice$)*P($\frac{Ace}{6\: on \: dice}$) reboot answered Jan 7, 2021 reboot comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes The answer should be P(Ace) = P(1on dice)*P(1Ace from deck) + P(2 on dice)*P(2 Ace from deck) + P(3 on dice)*P(3 Ace from deck) + P(4 on dice)*P(4 Ace from deck) rish1602 answered Sep 24, 2021 rish1602 comment Share Follow See 1 comment See all 1 1 comment reply rish1602 commented Nov 1, 2021 reply Follow Share Its 0.0816 0 votes 0 votes Please log in or register to add a comment.