If you draw a single card once from a deck of ordinary laying cards, what is the probability that it will be a 3 or a black card

The required probability =  0.5385

Step-by-step explanation:

The no. of cards in a deck = 52

The event of having  3 in a draw of a single card = 4/52

The event of having a black card in a draw of a single card = 26/52

The event of having  a black card and 3s = 2/52

Let T represent the event of having  3.

U to represent the event of having  a black card; &

V to represent the event of having  a black card that is 3s

Then the probability that it will be a 3 or a black card is;

P( T or U) = p(T) + p(U) - p(T and U)

P( T or U) = 4/52 + 26/52 - 2/52

P( T or U) = (4+26-2)/52

P( T or U) = 28/52

P( T or U)= 7/13

P( T or U) = 0.5385

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation: