statistics

The red face cards and the black cards numbered 2-9 are put into a bag. Four cards are drawn at random without replacement. Find the following probabilities:
a) All 4 cards are red
b) 2 cards are red and two cards are black
c) At least one of the red cards is red.
d) All four cards are black

  1. 👍
  2. 👎
  3. 👁
  1. a) There are 16 red-faced cards and 16 black.
    First Card red = 16/32
    Second card red = 15/31
    Third Card red = 14/30
    Fourth card red = 13/29
    The probability of all occurring is found by multiplying the individual probabilities.

    b) 16/32, 15/31, 16/30, 15/29 Again multiply.

    c) Firgure probability of one red card, two red cards, three red cards, or four red cards. Remember the probability of the remaning cards being black in each case, except the last. Since you are interested in "either-or", add those four probabilities.

    d) Same as a.

    1. 👍
    2. 👎
  2. There are 6 red cards not 16 ( only face cards: 2 kings 2 queens and 2 jacks) and 16 black. 22 in total.
    a) (6/22)*(5/21)*(4/20)*(3/19)
    b) (16/22)*(15/21)*(6/20)*(5/19)
    c) 1-((16/22)*(15/21)*(14/20)*(13/19))
    d)(16/22)*(15/21)*(14/20)*(13/19))

    1. 👍
    2. 👎
  3. CARDS:red=6; black=16 ... total 22

    a.(6 nCr 4)/(22 nCr 4) = .002
    b.(6 nCr 2)(16 nCr 2)/(22 nCr 4)= .246
    c.1-(.249) = .751 ... get d first
    d.(16 nCr 22)/(22 nCr 4) = .249

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. math

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

  2. statistic help please

    2. A survey of 259 families was made to determine their vacation habits. The two way table below sorts the families by location and the vacations by the length of the vacation. Rural Suburban Urban Total 1–7 days 80 39 37 156

  3. math

    Consider a standard deck of 52 playing cards with 4 suits. What is the probability of randomly drawing 1 card that is both a red card and a face card? (Remember that face cards are jacks, queens, and kings.) Enter your answer as a

  4. prealgebra

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

  1. Probability-Cards

    As shown above, a classic deck of cards is made up of 52 cards, 26 are black, 26 are red. Each color is split into two suits of 13 cards each (clubs and spades are black and hearts and diamonds are red). Each suit is split into 13

  2. Math

    Brent counted 10 red cards, 10 black cards, and 20 blue cards in a deck of cards. What is the ratio of red cards to the other cards?

  3. math

    A box contains 4 red cards, 6 blue cards, and 10 green cards. What is the smallest number of cards we must remove from the box to make sure that we have 2 cards of the same color?

  4. math

    A standard deck of cards has had all the face cards (Jacks, queens, and kings) removed so that only the ace through ten of each suit remains. A game is played in which two cards are drawn (without replacement) from this deck and a

  1. MATH Prob.

    You are dealt two cards successively without replacement from a standard deck of 52 playing cards. Find the probability that both cards are black.

  2. math

    Three cards are drawn at random without replacement from a standard deck of 52 playing cards. What is the probability that the second and third cards are black?

  3. Algebra

    How many hands of 5 cards can be made from a standard deck of cards such that each hand contains exactly 1 red card and 4 black cards?

  4. math

    Barry's deck of cards contains 40 blue cards and 70 red cards. Max's deck of cards contains the same number of blue cards, but the ratio of blue cards to red cards is 8:9. How many total cards does Max's deck of cards contain?

You can view more similar questions or ask a new question.