A Family Has Four Children. Use the Tree Diagram to Answer Each Question
A family has four children: Employ the tree diagram t0 answer each question:Choose the correct sample infinite {(F) , (M) } (FFFF), FFFM), (FFMF (FFMM); (FMFF (FMFM), (FMMF); (FMMM); (MFFF), (MFFM) , (MFMF) , (MFMM), (MMFF), (MMFM), (MMMF); (MMMM) {(FFFF), (FFFM), (FFMM), (FMMM), (MMMM) (FFFF), (FFFM); (FFMM) , (FMMM); (MFFF), (MMFF); (MMMF), (MmmM)0 ]
A family has four children: Utilize the tree diagram t0 answer each question: Choose the correct sample space {(F) , (1000) } (FFFF), FFFM), (FFMF (FFMM); (FMFF (FMFM), (FMMF); (FMMM); (MFFF), (MFFM) , (MFMF) , (MFMM), (MMFF), (MMFM), (MMMF); (MMMM) {(FFFF), (FFFM), (FFMM), (FMMM), (MMMM) (FFFF), (FFFM); (FFMM) , (FMMM); (MFFF), (MMFF); (MMMF), (MmmM) 0 ]
Figure 5-six shows histograms of several binomial distributions with n 5 6 trials. Match the given probability of success with the best graph. (a) p 5 0.30 goes with graph _________. (b) p v 0.50 goes with graph _________. (c) p 5 0.65 goes with graph _________. (d) p five 0.90 goes with graph _________. (due east) In full general, when the probability of success p is close to 0.5, would you say that the graph is more symmetric or more than skewed? In full general, when the probability of success p is close to ane, would you say that the graph is skewed to the correct or to the left? What about when p is shut to 0?
Since the probability off having a male child or daughter baby is equal. So property of each branch will be won by it. That is these branches equally lately. So now we are, you know, condition universe. Where they're given that at least i baby is a daughter. So except the one branch were 1st, 2nd and third Jailer boys, all other seven branches have a case. One girl merchandise instance. Initially, each branch is equally likely. And then in this condition universe too each branch will be equally legally and they're selling branches. So probability off each branch will be won past seven. And in that location is one branch were 1st, 2nd and 3rd Jailer girls. So one past seven is the probability that all three babies boots
And then the probability of having a male child or a girl is one-half. And the probability off each branch is when Wade So let f denote the issue Australia lives. Oh girl. Mhm. Exercise you do not live in all three drives? The girls? Yep, f corresponds to iv branches. It'south probability function four times one wait First half G responds to just one branch The probability of Gs one by eight and the jeep respond to just one branch hands Probability of ji intersection F is too in made. Then the provisional probability be off K unit. Yeah, be off Jean section F divided past be or fifth Yeah, What? Spun by iv.
And so probability of having a male child or a girl is half hence. Probability of each lunch is equal to 1 way. Elect G D non Lehman. All three childs, the girls, he did non. The event Third championship is oh, girl, Right. So they even be corresponds to four ranches. Hence probability of peas Four times when wade yous could accept jiggle response to one branch the probability of gs ane way. And since Yard correspond to just one branch hands probably orgy into section T is also unmade. And so the provisional probability P o chiliad yeah. Guarantee iss b f chiliad in the section t upon pft one by four
So probability of having a male child or a girl is one manner to Hence, the probability off each branch is ane style. Yeah. No, the grand d nor the event All three times The girls. Yeah, And asked, You lot know, the event Second child is a girl. You can so even this corresponds to four branches. Then probability of faces four times when Wade Bob G corresponds to one branch. So probability of gs i look instance K corresponds to one branch hands Probability of G intersection X is also ane by it. Then the conditional probability be off, 1000. Y'all and this. Yes, nosotros have g into sexiness. Exercise everybody? PFS sure is equal. Practise on my human foot.
v answers
La Due east, $ = 1,2,3be three*J element wy matrices corresporing to row row operations #five Bi ++ B+-3R, 6++ 8, due east = ? Ea ' R, +B Find the produrct B,B,E;coueh info lo uaVt Ur qustionnorc 0 thc outens
La E, $ = ane,two,3be iii*J element wy matrices corresporing to row row operations #5 Bi ++ B+-3R, 6++ eight, e = ? Ea ' R, +B Find the produrct B,B,East; coueh info lo uaVt Ur qustion norc 0 thc outens...
5 answers
Trouble 3. Consider a ii-factor modelYijl Oij + Eijl, i =one, Design a test ofK; j =1,1000, ane =1,Lij, Eijl are iid Due north (0,02).Ho 0ij = 0 for all i,j, against the culling that some of 0ij are dissimilar.
Problem 3. Consider a ii-factor model Yijl Oij + Eijl, i =1, Design a test of K; j =i, M, 1 =ane, Lij, Eijl are iid N (0,02). Ho 0ij = 0 for all i,j, against the alternative that some of 0ij are different....
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