Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Probability question for math please help out ?

Consider the following non-standard deck of cards. There are four colors (instead of the “suits” from a poker deck). For

each color there is a single card of each number 1,…,9. So there are 9 cards for each color and 36 cards in total.

a.) Suppose that Sammy wins if she draws a blue card. Otherwise, she loses.

What is the probability that she will win?

b.) Suppose now that Sammy wins if she draws a blue card, or she draws a seven. Otherwise she loses. In this case, what

is the probability that she wins?

c.) Jorge is going to draw one card from the same non-standard deck as above (Sammy has replaced her card, so all 36

cards are there). What is the probability that this card is neither a 5 nor a red card?

(Help with any or all) 

1 Answer

Relevance
  • 1 month ago
    Favourite answer

    a.) Suppose that Sammy wins if she draws a blue card. Otherwise, she loses.

    What is the probability that she will win?

    There are 9 ways to succeed; thus n(s = blue) = 9.  There are 27 ways to fail; so n(f) = 27.  The total number of possible outcomes is N = n(s) + n(f) = 36.  So by definition p(s = blue) = n(blue)/N = 9/36 = 1/4  ANS.

    b.) Suppose now that Sammy wins if she draws a blue card, or she draws a seven. Otherwise she loses. In this case, what is the probability that she wins? 

    She has n(s = blue) = 9 ways to win and n(s = 7) = 3 ways to win because one of the four sevens is a blue card (can't double count).  And the number of ways to fail n(f) = 27 - 3 = 24 ways because all 27 non-blue cards are failures except for the three 7's.

    And there you are p(s = blue OR = 7 not blue) = n(s = blue OR = 7)/36 = 12/36 = 1/3 ANS.

    c.) Jorge is going to draw one card from the same non-standard deck as above (Sammy has replaced her card, so all 36 cards are there). What is the probability that this card is neither a 5 nor a red card? 

    n(s <> 5) = 32 non fives in the pack; n(s <> red) = 27 - 24 = 3 as there are 8 non-fives in each of three non-red suits and we've already accounted for them.

    So p(s <> 5 OR s <> red) = n(s <> 5 OR s <> red)/N = 35/36 ANS.

    (Help with any or all) 

Still have questions? Get answers by asking now.