Solutions to Exercise on More on Probability.

  1. The solutions to this problem is evident when you have a tree diagram. This is a problem without replacement.
    1. The tree diagram should look like this.
    2. P(a black and a green) = P(black 1st and green 2nd) + P(green 1st and black 2nd)
      P(a black and a green) = (10/16)(6/15) + (6/16)(10/15)
      P(a black and a green) = (10/16)(6/15) + (6/16)(10/15) = 1/2
    3. P(black 1st and green 2nd)= (10/16)(6/15) = 1/4
    4. P(two same colours) = P(black 1st and black 2nd) + P(green 1st and green 2nd)
      P(two same colours) = (10/16)(9/15) + (6/16)(5/15) = 1/2

  2. The solutions to this problem is evident when you have a tree diagram. This is a problem without replacement.
    1. The tree diagram should look like this.
    2. P(a red and a green) = P(red 1st and green 2nd) + P(green 1st and red 2nd)
      P(a red and a green) = (5/10)(3/9) + (3/10)(5/9) = 1/3
    3. P(two same colours) = P(green and green) + P(blue and blue) +P(red and red)
      P(two same colours) = (3/10)(2/9) + (2/10)(1/9) + (5/10)(4/9) = 14/45

  3. A graphic way to solve this is to quicky sketch out a tree diagram as depicted here.
    1. This is basically picking a chocolate bar in the first pick. Thus, P(chocolate 1st) = 10/45 = 2/9
    2. The focus here is getting a granola in the second pick. There are two possibilities, Alfonso who picks first could pick a chocolate or a granola.
      P(granola 2nd) = P(chocolate 1st and granola 2nd)+ P(granola 1st and granola 2nd)
      P(granola 2nd) = (10/45)(35/44) + (35/45)(34/44) = 7/9
    3. P(both chocolate or granola)= P(both chocolate)+P(both granola)
      P(both chocolate or granola)= (10/45)(9/44) + (35/45)(34/44) = 64/99

  4. We could set up a tree diagram with 3 picks but it is easier to do the first questions by listing. Let ph, a, and n to denote peach, apple and nectarine respectively.
    1. P(ph, ph, ph) = (5/15)(4/14)(3/13) = 2/91
    2. P(first three all have different fruits) = P(ph,a,n) + P(ph,n,a) + P(a,n,ph) +P(a,ph,n) + P(n,a,ph) + P(n,ph,a)
        = (5/15)(4/14)(6/13) + (5/15)(6/14)(4/13) + (4/15)(6/14)(5/13) + (4/15)(5/14)(6/13) + (6/15)(4/14)(5/13) + (6/15)(5/14)(4/13)
        = 6[5 x 4 x 6]/[15 x 14 x 13] = 24/91
    3. Here it might be easier to have a tree diagram up to the second pick. Here we want to find P(peach in the second pick). There are three possibilities. P(peach in the second pick) = P(ph, ph) + P(a, ph) + P(n, ph)
      p(peach in the second pick) =(5/15)(4/14) + (4/15)(5/14) + (6/15)(5/14)
      p(peach in the second pick) = 1/3

    1. Each black square denotes a possible combination. The diagonal of the table is empty because the two students cannot be the same person. The bottom half of the diagonal is also empty because (A,B) and (B,A) are the same combination because order of selection does not matter. After all both (A,B) and (B,A) say that Albert and Bertha get selected.
      Total possible cominations = 15.
    2. P(Devi) = 5/15 = 1/3 from counting black boxes that involve D.
    3. P(Devi and Enrique) = 1/15 from the table.

  5. A class has 50 student and each student either plays or not play the violin.
    1. First calculate the total number of male students. Then calculate the total number of female students. Next, find the number of female students who do not play violin.
        plays violin does not play violin Total
      Male 16 10 10+16= 26
      Female 24-12=12 12 50-26=24
      Total 16+12=28 10+12=22 50
    2. P(female and plays violin)= 12/50 from the relevant cell in the table.
    3. This is a difficult question. P(both female or both play violin) = P(both female) + P(both play violin) - P(both female and play violin)
      Note that p(both female and play violin) is here to avoid double accounting. This is a problem of without replacement. Thus, P(both female) = (24/50)(23/49) because the probability of obtaining the second female after the first female student has been selected is 23 out of 49 students. Similarly p(both female AND play violin) = (12/50)(11/49). The P(a female and play violin) =12/50 from the relevant cell on the table. But the p(second female and play violin) is 11 females out of 49 students.
      P(both female or both play violin) = P(both female) + P(both play violin) - P(both female and play violin)
      P(both female or both play violin) =(24/50)(23/49)+ (28/50)(27/49) - (12/50)(11/49)
      P(both female or both play violin) = 12/25