In a Binomial Distribution consisting of 4 independent trials, probabilities of 2 and 3 successes are 0.35986 and 0.17993 respectively. Find the probability of success. 09:162/2Normal Distribution Table0.050.060.010.00399 0.00798 0.01197 0.01595 0.01994 0.02392 0.02790.020.030.040.070.080.090.03188 0.035860.04776 0.05172 0.05567| 0.05962 0.06356 0.06749 0.07142 0.075350.03983 0.04380.07926 0.08317 0.08706 0.09095 0.09483 0.09871 0.10257 0.10642 0.11026 0.114090.3 0.11791 0.12172 0.12552 0.12930.4 0.15542 0.1591 0.16276 0.16640.5 0.19146 0.19497 0.19847 0.20194 0.20540.22575 0.22907 0.23237 0.23565 0.23891 0.24215 0.24537 0.24857 0.25175 0.25490.25804 0.26115 0.26424 0.26730.28814 0.29103 0.29389 0.29673 0.29955 0.30234 0.30511 0.30785 0.31057 0.313270.31594 0.31859 0.32121 0.323810.34134 0.34375 0.34614| 0.34849 0.35083 0.35314 0.35543 0.35769 0.35993 0.362140.10.20.13307 0.13683 0.14058 0.14431 0.14803 0.151730.17003 0.17364 0.17724 0.18082 0.18439 0.187930.20884 0.21226 0.21566 0.21904 0.22240.60.70.27035 0.27337 0.27637| 0.27935 0.28230.285240.80.90.32639 0.32894 0.33147 0.33398 0.33646 0.3389110.36864 0.37076 0.37286 0.37493 0.37698 0.3790.3810.382981.2 0.38493 0.38686 0.38877 0.39065 0.39251 0.39435 0.39617 0.39796 0.39973 0.401471.3 0.40321.4 0.41924 0.42073 0.42221.50.417740.42364 0.42507 0.42647 0.42785 0.42922 0.43056 0.431890.40490.40658 0.40824 0.40988 0.41149 0.41308 0.41466 0.416210.43319 0.43448 0.43574 0.43699 0.43822 0.43943 0.44062 0.44179 0.44295 0.444080.45053 0.45154 0.45254 0.45352 0.454490.46164 0.46246 0.463271.60.44520.44630.44738 0.44845 0.44950.45543 0.45637 0.45728 0.45818 0.45907 0.45994 0.46080.46407 0.46485 0.46562 0.46638 0.46712 0.46784 0.46856 0.46926 0.46995 0.470621.9 0.47128 0.47193 0.47257 0.47320.47725 0.47778 0.47831| 0.47882 0.47932 0.47982 0.48030.48214 0.48257 0.4832.2 0.48610.48928 0.48956 0.48983 0.49012.4 0.49182.5 0.49379 0.49396 0.49413 0.49432.6 0 0.49573 0.49585 0.49598 0.49609 0.49621 0.49632 0.496431.71.80.47381 0.47441 | 0.4750.47558 0.47615 0.476720.48341 0.48382 0.48422 0.48461 0.4850.48645 0.48679 0.48713 0.48745 0.48778 0.48809 0.48842.10.48537 0.485740.48870.488990.49036 0.49061 0.49086 0.49111 0.49134 0.491580.49202 0.49224 0.49245 0.49266 0.49286 0.49305 0.49324 0.49343 0.493610.49446 0.49461 0.49477 0.49492 0.49506 0.49522.30.49534 0.49547 0.49560.49653 0.49664 0.49674 0.49683 0.49693 0.49702 0.49711 0.49720.49744 0.49752 0.4976 0.49767 0.49774 0.49781 0.49788 0.49795 0.49801 0.498070.49813 0.49819 0.49825 0.498310.49865 0.49869 0.49874| 0.49878 0.49882 0.49886 0.49889 0.49893 0.49896 0.4993.1 0.49903 0.49906 0.49910.49931 0.49934 0.49936 0.49938 0.49940.49952 0.49953 0.49955 0.49957 0.49958 0.49960.49966 0.49968 0.49969 0.49970.49977 0.49978 0.49978 0.49979 0.49980.49984 0.49985 0.49985 0.49986 0.49986 0.49987 0.49987 0.49988 0.499883.7 0.49989 0.49993.82.70.49728 0.497362.82.90.49836 0.49841 0.49846 0.49851 0.49856 0.4986130.49913 0.49916 0.49918 0.49921 0.49924 0.49926 0.499290.49942 0.49944 0.49946 0.49948 0.49950.49961 0.49962 0.499640.499650.49971 0.49972 0.49973 0.49974 0.49975 0.499760.49981 0.49981 0.49982 0.49983 0.499833.23.33.43.53.60.49990.49991 0.49991 0.49992 0.49992 0.49992 0.499920.49990.49993 0.49993 0.49993 0.49994 0.49994 0.49994 0.49994 0.49995 0.49995 0.499950.49995 0.49995 0.49996 0.49996 0.49996 0.49996 0.49996 0.49996 0.49997 0.499970.49997 0.49997 0.49997| 0.49997 0.49997 0.49997 0.49998 0.49998 0.49998 0.499983.94A uzem.altinbas.edu.tr

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Answered: In a Binomial Distribution consisting…

In a Binomial Distribution consisting of 4 independent trials, probabilities of 2 and 3 successes are 0.35986 and 0.17993 respectively. Find the probability of success.

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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