T 0.1 0.2 0.3 0 0.9048 0.8187 0.7408 1 0.9953 0.9825 0.9998 1.0000 2 3 4 5 6 T 3 4 5 6 7 8 9 10 T 1.0 1.5 0 0.3679 0.2231 1 0.7358 2 0.9197 11 12 13 14 15 16 0.1 T Poisson Probability Sums p(x;µ) 0.9999 1.0000 0.9989 0.9964 0.9999 0.9997 1.0000 1.0000 0.2 0.5578 0.8088 0.9810 0.9344 1.0 0.9963 0.9814 0.9994 0.4 0.6 0.9 0.6065 0.5488 0.4066 0.7725 0.6703 0.9631 0.9384 0.9098 0.8781 0.8442 0.9921 0.9856 0.9769 0.9659 0.9992 0.9982 0.9966 0.9996 0.9526 0.9371 0.9942 0.9909 0.9865 0.9999 0.9998 0.9992 0.9986 0.9977 1.0000 1.0000 1.0000 0.9999 0.9998 0.9997 1.0000 1.0000 1.0000 0.7 0.8 0.9 0.3 1.5 0.9473 0.9955 0.9834 2=0 2.0 2.5 0.1353 0.0821 0.4060 0.2873 0.6767 0.5438 0.8571 0.7576 0.4 2.0 " 0.5 0.5 " f 3.0 0.9991 0.9955 0.9858 0.9665 0.9998 0.9989 1.0000 0.0498 0.1991 0.4232 0.6472 0.8912 0.8153 0.9580 0.9161 2.5 1.0000 0.9999 1.0000 0.6 0.9347 0.8893 0.8311 0.9733 0.9489 0.9134 0.9958 0.9881 0.9998 0.9989 0.9962 0.9901 1.0000 0.9997 0.9989 0.9967 0.9786 0.9597 0.9919 0.9829 0.9999 0.9997 0.9933 3.0 " 0.7 0.8 0.4966 0.4493 0.8088 3.5 4.0 4.5 5.0 0.0302 0.0183 0.0111 0.0067 0.1359 0.0916 0.0611 0.0404 0.3208 0.2381 0.1736 0.1247 0.5366 0.4335 0.3423 0.2650 0.7254 0.6288 0.5321 0.8576 0.7851 0.7029 0.9990 0.9972 0.9991 0.9976 0.9997 0.9999 0.9997 0.9992 1.0000 0.9999 0.9997 1.0000 0.9999 1.0000 3.5 4.0 4.5 0.4405 0.6160 0.7622 0.8666 0.9319 0.9682 0.9863 0.9945 0.9980 0.9993 0.9998 0.9999 1.0000 5.0 On average, 2.5 traffic accidents per month occur at a certain intersection. Complete parts (a) through (c) below. Click here to view the table of Poisson probability sums. (a) What is the probability that exactly 2 accidents will occur in any given month at this intersection? The probability that exactly 2 accidents will occur in any given month at this intersection is 0.2565 (Round to four decimal places as needed.) (b) What is the probability that fewer than 5 accidents will occur in any given month at this intersection? The probability that fewer than 5 accidents will occur in any given month at this intersection is (Round to four decimal places as needed.)

Part D?

1 1 and· 5'6 1. The probabilities of three contestants X,Y, and Z of winning the grand prize i 3 respectively. Assuming that only one contestant can win the grand prize, find the probability that a. Either X or Y wins 1 11 Answer: P (X or Y) = P(X) + P (Y) =+ = 1 30 b. neither X nor Z wins

Can you explan it better?  I do not understand.