A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ( and , called `hearts' and `diamonds') are red, the other two ( and , 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

The probability that the first two cards drawn from the deck are both red is 24.51%

Step-by-step explanation:

A standard deck of cards has 52 cards. There are 4 of each card (4 Aces, 4 Kings, 4 Queens, etc.) . There are 4 suits (Clubs, Hearts, Diamonds, and Spades) with each suit being 13 cards .  Two of the suits (hearts and diamonds) are red, the other two (spades and clubs) are black.

P(first two cards drawn from the deck are both red)= P(First card is red) ∙× P(Second card is red)

Since two suits of the card is red (i.e hearts and diamond) and each suit consist of 13 cards, the total number of red cad = 13 + 13 = 26

P(First card is red) = number of red card / total number of cards = 26 / 52

When the first card is drawn without replacement i.e the card is not put back into the deck, the total number of card reduces to 51 and the number of red card reduces to 25.

Therefore, P(Second card is red) = 25 / 51

P(first two cards drawn from the deck are both red)= P(First card is red) ∙× P(Second card is red) = (26 / 52) × (25 / 51) = 0.2451 = 24.51%

The probability that the first two cards drawn from the deck are both red is 24.51%

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)

cant open file sorry!

step-by-step explanation: