Assume that a procedure yields a binomial distribution with n=5 trials and a pro

Question

Assume that a procedure yields a binomial distribution with n=5 trials and a probability of success of p=0.6

Binomial Probabilities 6 n 8 NOTE: 0 represents a positive probability less than 0.0005. 2 0+ 0.002 0.0 0.04 0.09 0.16 0.25 0.36 0.49 0.64 0.81 0.902 0.98 0 097 857 0.729 0512 0.216 0.125 0.064 0.027 0.008 0.0010 0 3 0.029 0,135 0,243 0.384 0,441 0.432 0.375 0,288 0.189 0,096 0,027 0,007 04 2 0+ 0.007 0.027 0.096 0.189 0.288 0.375 0.432 0.441 0.384 0.243 0.135 0.029 30 0 0.001 0.008 0.027 0.064 0.125 0.216 0.343 0.512 0729 0.857 0.97 3 0.961 0,815 0.656 0.41 0.24 0.13 0,062 0.026 0.008 0,002 04 1 (0.048 2 0001 0.021 0.073 0205 o 0346 0312 1023 0132 0051 0008 00010+ 2 0.057 2 0001 0.031 Oog8 0246 0.324 0311 (0234 0138 006 0015 00010 0+ 2 0.311 LU 0.001 0.004 0.016 0.047 0.118 0.262 0.531 0.735 0.941 o 0.932 0.698 0.478 0.21 0.082 0.028 0.008 0.002 0 0 0+ o 0 7 0.066 0,257 0,372 0.367 0,247 0,131 0,055 0.01 0.004 0+ 2 0002 0.041 0124 0275 0261 0164 0.077 0.025 0004 0 0+ 2 0.194 0.097 0.232

Explanation / Answer

If we will see in the column of p = 0.6 and in the row corresponding to n = 5 and x = 3, the probability is 0.346

Hence,

P(Exactly 3) = 0.346

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