Assume that a procedure yields a binomial distribution with nequals=4 trials and

Question

Assume that a procedure yields a binomial distribution with

nequals=4 trials and a probability of success of p=0.8

20 0,002 001 009 016 o 25 36 0,49 0.64 081 0902 098 20 0007 0.027 0119 0288 0.975 0441 0.384 0243 0135 0029 2 4 0 0961 815 0.656 024 013 0002 0026 0,008 000 0 4 0.15 1 0048 0204 0528 036 0259 0.156 0.07 0.028 0.006 0 30 0,001 0008 0.051 0 132 023 0512 0346 0 309 205 0.073 0021 0001 951 0104 30 0002 0.015 0082 0185 0276 0912 0276 0185 0.082 0.015 0002 50 0 0002 001 0.037 0.094 0187 0.303 0593 0.354 0232 0057 5 60 0 0 0 0,001 0,004 0016 0118 0.262 0531 0735 0941 6 0 932 0.698 0478 021 0,082 0.028 0.008 0.002 0 0 0 7 0.0040+ 60 0 0004 0,017 005 01 31 0247 0567 0372 0257 0066 6 0 0923 663 043 0168 0,058 0.017 0.004 0.001 0 0 0 0 1 0075 0279 0583 0198 009 00 31 0008 00010 0 0 1 40 0.005 0045 0136 0232 0213 0232 0156 0.046 0.005 0 4 047 60 0 0001 001 0,041 109 0.296 0.294 0149 0051 0003 6 n x 0.05 0.1 02 04 05 0.6 07 08 09 095 099 x n NOTE: 0+ represents a positive probability less than 0.0005

Explanation / Answer

Solution:- The probability that the number of successes x is exactly 3 is 0.4096.

p = 0.80, n = 4, x = 3

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

P(x, n, p) = 4C3*(0.8)3 *(1 - 0.8)(4 - 3)

P(x = 3) = 0.4096

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