Eric is randomly drawing cards from a deck of 52. He first draws a red card, places it back in the deck, shuffles the deck, and then draws another card. What is the probability of drawing a red card, placing it back in the deck, and drawing another red card

Respuesta :

Half the deck is red so you have. 1/2 chance of picking red each time.

Picking red both times would be 1/2 x 1/2 = 1/4

The answer is 1/4

The  probability of drawing a red card, placing it back in the deck, and drawing another red card is 0.25 or [tex]\bold{\frac{1}{4}}[/tex].

Given to us:

Eric is randomly drawing cards from a deck of 52.

Also,  Eric first draws a red card, places it back in the deck, meaning this is case of probability with replacement (outcomes are returned back to the sample space again).

So, the both the events are independent of each other, therefore the probability for second event will be same as the first, as the sample size and the desired outcome are same.

Probability for first event

                  [tex]=\dfrac{Desired\ outcome}{Total\ number\ of\ samples}\\\\=\dfrac{26}{52} \\\\=\dfrac{1}{2} = 0.5[/tex]

The probability of drawing a red card, placing it back in the deck, and drawing another red card

= Probability for first event x Probability for second event

= 0.5 x 0.5

= 0.25

Hence, the  probability of drawing a red card, placing it back in the deck, and drawing another red card is 0.25 or [tex]\bold{\frac{1}{4}}[/tex].

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