Nutwin Posted November 19, 2007 Report Share Posted November 19, 2007 Hello,I have a question about probability which i can't seem to figure out, here it is:50 students go bushwalking. 23 get sunburnt, 22 get bitten by ants and 5 are both sunburnt and bitten by ants.A child is selected at random. Determine the probability that the child has:a) escaped being bittenB) was either bitten or sunburntc) was neither bitten nor sunburntd) was bitten, given that the student was sunburnte) was sunburnt, given that the student was not bitten.I would really appreciate it if someone could explain this to me because I've been stuck on this for 3 hours and it's driving me crazy.Thank you! Link to post Share on other sites More sharing options...
Abu Posted November 20, 2007 Report Share Posted November 20, 2007 That's easy. To make the Venn Diagram, you'll have 18 as a, 5 as b, 17 as c and 10 as d.So escaped bitten would be 22/50 (17+5) 17 being just bitten, and 5 being both bitten and sunburnt.either bitten or sunburnt is 5/50 or 1/10neither would be 10/50, so 1/5d) 5/23, since there were 23 people(18+5) that got sunburnte) 23/33, since 23 were sunburnt, and 50 - 17 students that were bitten. Link to post Share on other sites More sharing options...
Guest tbittner Posted November 20, 2007 Report Share Posted November 20, 2007 (edited) Hi-Here's how you go about it....You know that there are 50 students altogether. Let S = the event that a student gets sunburnt, Let A = the event that a student gets bitten by ants. Therefore, we are given the following information:P(S) = 23/50, P(A) = 27/50, P(S AND A) = 5/50From this, we know that 23 - 5 = 18 students only got sunburnt, and 27 - 5 = 22 students only got bitten.a) P(student escapes being bitten) = P(not A) = 23/50b. P(A OR S) = P(A) + P(S) – P(A AND S) = 27/50 + 23/50 – 5/50 = 45/50 = 9/10 c) P(neither bitten nor sunburnt) = 1 - 9/10 = 1/10 (the opposite of part b )d) P(A given S) = P(A AND S)/P(S) = (5/50)/(23/50) = 5/23 (Bayes' Rule)e) P(S given not A) = P(S AND NOT A)/P(NOT A) = (18/50)/(23/50) = 18/23 You can draw a Venn diagram of this problem by putting 18 in the circle containing only S, 22 in the circle containing only A, and then 5 in the overlapping part of the circle. Since 18 + 22 + 5 = 45, there are 5 students who are not in the circles at all.Keep in mind that sometimes these problems are worded strangely, and poorly worded problems can mean that the answer the teacher has in mind might be different. However, given the way this problem is worded, this is the answer. Hope this helps!Hello,I have a question about probability which i can't seem to figure out, here it is:50 students go bushwalking. 23 get sunburnt, 22 get bitten by ants and 5 are both sunburnt and bitten by ants.A child is selected at random. Determine the probability that the child has:a) escaped being bittenB) was either bitten or sunburntc) was neither bitten nor sunburntd) was bitten, given that the student was sunburnte) was sunburnt, given that the student was not bitten.I would really appreciate it if someone could explain this to me because I've been stuck on this for 3 hours and it's driving me crazy.Thank you! Edited November 20, 2007 by tbittner Link to post Share on other sites More sharing options...
Nutwin Posted November 22, 2007 Author Report Share Posted November 22, 2007 Hey thanks so much for helping, both of you! I really appreciate it! I hope i passed my test Link to post Share on other sites More sharing options...
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