Can someone help me with this question?
2007-07-08 14:10:46 · 5 answers · asked by Mega 1 in Science & Mathematics ➔ Mathematics
There are 4 queens and 4 kings in the deck, so p = 8/52 = 0.154
2007-07-08 14:13:22 · answer #1 · answered by TychaBrahe 7 · 0⤊ 0⤋
I assume you're looking for probability.
In a 52 card deck, there are four suits, and 1 king and 1 queen for each suit. Therefore, there are 4 kings and 4 queens (a total of 8) in a deck.
8/52=2/13 chances to draw a queen or a king.
Is that what you're looking for?
2007-07-08 14:15:09 · answer #2 · answered by Shauna 3 · 0⤊ 0⤋
4 queens out of the 52-card deck
4 kings out of the 52-card deck
the probability to draw a queen is 4/52
the probability to draw a king is 4/52 there for the probability to ba a queen or a king= 4/52+4/52=8/52
p(queen or king)=2/13
2007-07-08 14:24:19 · answer #3 · answered by sousouch 2 · 0⤊ 0⤋
there are 4 queens and 4 kings. A card is either a queen or a king, hence the events queen and king are disjoint,
so P(Q U K) = 8/52
2007-07-08 14:14:13 · answer #4 · answered by swd 6 · 0⤊ 0⤋
there are 4 ace enjoying cards in a deck of enjoying cards probabilty of having an ace is = 4/fifty two there are 13 enjoying cards in a deck that are golf equipment yet u choose the risk of the cardboard no longer being a club no of enjoying cards which isn't a club is= fifty two-13 = 39 danger of the cardboard no longer being a club=39/fifty two combining the above 2 we get that........ the risk of the cardboard being an ace or no longer a club is = 4/fifty two + 39/fifty two -one million/fifty two i'm doing -one million/fifty two as there is an ace card that's a club and u dont choose to calculate the risk for it two times the final answer is= 40 two/fifty two
2016-12-14 03:07:47 · answer #5 · answered by ? 4 · 0⤊ 0⤋