Welcome to our Homework Page for the Yellow Group in P2b!
Please submit homework by Monday and ensure that your child brings their homework folder with reading book each day to read aloud in their group.
Welcome back to a new term!
N.B. If you have not received your date and time for the Parent Consultations please contact the school as soon as possible – these have been handed out.
Welcome back to everyone after a well deserved October break.
We have a newHomework Grid starting this term. Paper copies are in the homework jotters but there is an electronic copy underneath this week’s main homework.
Please continue to work through your Numeracy Grids as usual.
P2 Homework Grid October to December
03.10:Christmas card templates will be going home this week to be returned byThursday 13th October. For best results, stay within the border and complete using felt tip pens, coloured pencils, wax crayons or paint. Please DO NOT add stickers, glitter, metallic paper or 3D objects.
Math 431 – Fall 2015 Homework 4 – Solutions 2.15) We have an urn with three green balls and two yellow balls. We pick a sample of two without replacement and put these two balls in a second urn that was previously empty. Next we sample two balls from the second urn with replacement . (a) What is the probability that the ﬁrst sample had two balls of the same color? Solution: Let A 1 be the event that the ﬁrst sample had two balls of the same color. Note that we ﬁrst sample without replacement, so e.g. we consider it with order then there are 5 · 4 possible outcomes. Counting the green-green and yellow-yellow cases separately we get that 3 · 2 + 2 · 1 of those outcomes will have two balls of the same color. Thus P ( A 1 ) = 3 · 2 + 2 · 1 5 · 4 = 2 5 . (You would get the same answer if you took the sample without order.) (b) What is the probability that the second sample had two balls of the same color? Solution: Let A 2 be the event that the second sample had two balls of the same color. We have P ( A 2 | A 1 ) = 1, since if the ﬁrst sample had two balls of the same color then this must be true for the second one. We also have P ( A 2 | A c 1 ) = 1 2 , because if we have to sample from an urn containing a yellow and a green ball twice with replacement then 1 / 2 would be the probability that the second outcome has the same color as the ﬁrst one. We already know P ( A 1 ) = 2 5 and P ( A c 1 ) = 3 5 which gives P ( A 2 ) = P ( A 2 | A 1 ) P ( A 1 ) + P ( A 2 | A c 1 ) P ( A 1 ) c = 1 · 2 5 + 1 2 · 3 5 = 7 10 . (c) Given that the last two balls have the same color, what is the probability that the second urn contains two balls of the same color? Solution: Using the already computed probabilities: P ( A 1 | A 2 ) = P ( A 2 | A 1 ) P ( A 1 ) P ( A 2 ) = 1 · 2 5 7 10 = 4 7 . 2.28) Show that if P ( A ) = 0 or P ( A ) = 1 then any event B is independent of A . Solution: Suppose that P ( A ) = 0. Since for any event B we have AB ⊂ A and P ( AB ) ≤ P ( A ) = 0 we also get P ( AB ) = 0. Then 0 = P ( AB ) = P ( A ) P ( B ) = 0 · P ( B ) = 0 which shows that A and B are independent. (Note that we cannot assume that A is the empty set!) Now suppose that P ( A ) = 1. Then P ( A c ) = 0 and by the ﬁrst part A c will be independent of any event B . But if A c and B are independent then the same is true for A and B . 2.30) We have a system that has 2 independent components. Both components must function in order for the system to function. The ﬁrst component has 8 independent elements