Answered

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In a bag, there are 5 cards numbered 1 to 5. Each time a card is randomly selected, it I'd replaced in the bag. What is the probability Marsha will select an even-numbered card first and then an odd-numbered card?

Sagot :

ChillinFool
there is a 2 out of 5 chance of pulling an even[tex] \frac{2}{5} [/tex]
there is a 3 out of 5 chance of pulling an odd the second time[tex] \frac{3}{5} [/tex]

[tex] \frac{2}{5} [/tex]*[tex] \frac{3}{5} [/tex]=[tex] \frac{6}{25} [/tex]

there is a 6 out of 25 chance of it happening in that order
AL2006

Out of 5 cards in the bag ...

-- There are 2 even numbers . . . 2 and 4 .

-- There are 3 odd numbers . . . 1, 3, and 5 .

So EITHER time, the probability of pulling an even number is  2/5 ,
and the probability of pulling an odd number is  3/5 .

The probability of pulling an even number the first time
AND an odd number the second time is

       (2/5) x (3/5)  =  6/25  =  24% .

 
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