Please help my Math homework

asked Dec 27, 2016 in Other Math Topics by anonymous

You need to submit fewer questions at once. There are limitations in the system restricting the length of answers. This site isn't here to do your homework for you. It's here to help you tackle problems for yourself. Please don't use it as a quick fix! Answers are not guaranteed to be correct.

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:

To avoid this verification in future, please log in or register.

1 Answer

8) Of the 25 that like football, 10 like hockey as well so only 2 like hockey but not football, and 15 like football but not hockey. So we have 15+10+2=27 students out of 40 that like football or hockey or both. Also, 15+2=17 like football or hockey, but not both, so the probability (a) of selecting a student who likes one or the other (but not both) is 17/40=0.425 or 42.5%.

(b) 50-27=23 students like neither game, so that's 23/40=0.575 or 57.5% (this is 1 or 100%-(a)).

(c) Venn diagram:

Answers to follow in comments...

answered Dec 28, 2016 by Rod Top Rated User (425,260 points)


Tree diagram above.

Probability of all failing is 0.056.

Probability of at least one pass is the complement of this = 1-0.056=0.944.

The 8 outcomes (top to bottom), D=Dave, M=Mike, K=Ken, P=pass, F= fail:

DP,MP,KP (All pass=0.144)





DF,MP,KF (Only Mike passes=0.084)


DF,MF,KF (All fail=0.056)

To find the probability, multiply together the probabilities encountered on the path to the "leaf". You can then fill in the gaps for the outcomes,


  1 2 3 4 5 6
HEAD H1 H2 H3 H4 H5 H6
TAIL T1 T2 T3 T4 T5 T6

6 of these possible outcomes contain a head, but we have to remove H3 because this event is head AND 3 and we want head OR 3. So there 6 outcomes: probability is 6/12=1/2 (H1 H2 H4 H5 H6 T3).

C1) There are 18 fruits in all.

a) 0, impossibility because there are no durians.

b) 8/18=4/9, because there are 8 apples.

c) 1-4/9=5/9, because this is the complement of (b).


The complete circle on the left represents P(A) and the one on the right P(B). X, Y and Z are regions such that:

P(A)=X+Y=0.5; P(A^B)=Y=0.35; P(AvB)=X+Y+Z=0.75.

Since Y=0.35, X=0.5-0.35=0.15 and Z=0.25

So P(B)=Y+Z=0.35+0.25=0.6.


a) 6 outcomes out of 15 possible outcomes result in a green marble.

b) 5 outcomes out of 15 possible outcomes result in a blue marble.


a) 1/6

b) 3/6=1/2 because there are 3 odd numbers: 1, 3, 5


a) 3/6=1/2 because there are 3 primes: 11 13 17

b) also 1/2 because 15 21 51 are divisible by 3


a) 12/57=4/19,

b) 25/57, because there 57 stickers in all


a), c), d) 3/6=1/2 because there are 3 odd and 3 even numbers.

b) 4/6=2/3 because 1, 2, 3, 4 make 4 numbers less than 5.

e) An odd number is 1 3 5, a number less than 5 is 1 2 3 4. We have to eliminate elements that belong to both sets, so we end up with 2 4 5. Each one of these is EITHER an odd number OR less than 5. It isn't BOTH. So it satisfies the conditions. There are 3 numbers out of a possible 6, so the probability is 3/6=1/2 again.


This diagram has been copied from 8c. The regions have been renamed.

Let's see how the figures were obtained. Region abbreviations: H=married male (husband), W=married female (wife), S=unmarried female (spinster), B=unmarried male (bachelor).

B+H=Male=381; H=168, so B=381-168=213.

Married=H+W=299, so W=299-168=131.

(Incidentally, S+B+H+W=655 people with multiple jobs, so S=143.)

We must only have unmarried males or married females to satisfy the conditions, so that's B+W=213+131=344. Therefore 344 people are EITHER married OR male, but NOT BOTH. That means we count bachelors and wives, and exclude husbands and spinsters. The probability of selecting a male or married person is therefore 344/655=0.5252 approx.

5) Use the diagram in C2. From the given figures we have 4 pieces of information:

  1. The probability of getting grade D only (implies a pass)=1/6
  2. The probability of not getting grade D (implies fail, or passing with a grade better than D)=5/6
  3. The probability of failing=1/9
  4. The probability of passing=8/9

Now let's define the regions X, Y, Z.

X probability of failing=1/9 (assuming that any grade below D is a fail; see (3) above)

Y probability of getting grade D=1/6 (see (1) above)

Z probability of getting better than grade D (we have to find)

X+Y=2/18+3/18=5/18, left circle (probability of failing or getting grade D)

Y+Z=probability of passing=8/9, right circle (see (4) above) so Z=8/9-1/6=13/18

X+Y+Z must equal 1 because it is certain that a student will pass or fail. And 1/9+1/6+13/18 does add up to 1. Also, (2) above is X+Z=1/9+13/18=15/18=5/6, so this fits, too.

Probability of getting a better grade than D is Z=13/18.

Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
77,983 questions
81,731 answers
60,939 users