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March 29, 2017

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1)Two marbles are chosen at random from a bag containing 3 blue and 2 red marbles. The relative-frequency histogram shows the distribution of the number of red marbles chosen. Find P(0 red). (the gragh shows 3/10 for 0 red, 3/5 for 1 red, and 1/10 for 2 red)

A)0
B)8/15
C)3/10
D)2/5
I chose C

2)A red die and a blue die are tossed. What is the probability that the red die shows a 3 and the blue die shows a number greater than 3?

A)1/10
B)1/5
C)1/12
D)3/5
I chose C

3)Tickets are numbered 1 to 50 and placed in a box. Three tickets are drawn at random without replacement. What is the probability that their numbers are all greater than 25?

A)1/8
B)23/196
C)69/625
D)1/2
I chose D

4)From 4 yellow and 8 blue marbles, 3 are selected. What is the probability that all 3 are yellow or all 3 are blue.

A)3/11
B)1/55
C)14/55
D)3/220
I don't know I GUESS A

5)A card is drawn from a standard deck of cards. What is P(heart or a 6)?

A)9/26
B)17/52
C)1/4
D)4/13
I chose B

  • Probability (#5 of 5) - ,

    5) There are 13 hearts and 3 sixes that are not hearts. You can't count the six of hearts twice. That makes 16/52. Which fraction choice is that?

  • Probability/Statistics - ,

    1 and 2 are correct

    for 3 I got 23/196

    there are 25 tickets greater than 25
    so prob = 25/50 x 24/49 x 23/48 = 23/196

    for 4 you have to use a formula that says:
    P(A or B) =P(A) + P(B) - P(A and B)

    since the prob of them being 3 yellow AND 3 blue is zero

    P(3yellow) = 4/12 x 3/11 x 2/10 = 1/55
    P(3blue) = 8/12 x 7/11 x 6/10 = 14/55

    so your prob = 1/55 + 14/55 = 3/11

    GOOD GUESS

    #5 (use the same formula as #4)

    prob(heart) = 13/52
    Prob(6) = 4/52
    Prob(6of hearts) = 1/52

    so prob(heart OR a 6) = 13/52 + 4/52 - 1/52
    = 16/52
    = 4/13

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