Asked by Help
A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ($\heartsuit$ and $\diamondsuit$, called `hearts' and `diamonds') are red, the other two ($\spadesuit$ and $\clubsuit$, called `spades' and `clubs') are black. The cards in the deck are placed in random order (usually by a process called `shuffling'). What is the probability that the first two cards drawn from the deck are both red?
Answers
Answered by
CodyJinks
(26/52)+(25/51)
Answered by
Reiny
No, the second event is dependent on the first, so....
(26/52)(25/51) = .....
(26/52)(25/51) = .....
Answered by
Damon
I bet Cody meant * not +
Answered by
Helen
It is (26/52)(25/51) Which is equal to 25/102
Answered by
wut
its not correct
Answered by
-A
The correct answer is 25/102.
There are 26 red cards in the set, and 52 cards.
The first card has a
$1/2$ chance of being red and this question is without replacement, meaning it has a dependant equation, and so the next fraction in the equation will be $25/51$.
Now, we multiply $1/2$ and $25/51$, and get $25/102$.
There are 26 red cards in the set, and 52 cards.
The first card has a
$1/2$ chance of being red and this question is without replacement, meaning it has a dependant equation, and so the next fraction in the equation will be $25/51$.
Now, we multiply $1/2$ and $25/51$, and get $25/102$.
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