= Solution
Under the null hypothesis, the defect count in each packet has a <binomial distribution> with parameters $3,\theta$. The <likelihood>, apart from factors independent of $\theta$, is
$$
L(\theta)\propto\theta^{94+2(40)+3(6)}(1-\theta)^{3(116)+2(94)+40}=\theta^{192}(1-\theta)^{576}.
$$
The <maximum-likelihood estimate> is therefore
$$
\boxed{\widehat\theta=\frac{192}{768}=\frac14.}
$$
The fitted <binomial distribution> probabilities for counts $0,1,2,3$ are $27/64,27/64,9/64,1/64$, yielding expected packet counts $108,108,36,4$. The <Pearson chi-squared goodness-of-fit test> uses
$$
X^2=\frac{(116-108)^2}{108}+\frac{(94-108)^2}{108}+\frac{(40-36)^2}{36}+\frac{(6-4)^2}{4}=\boxed{\frac{104}{27}\simeq3.852}.
$$
For <binomial goodness-of-fit with an estimated parameter>, there are $4-1-1=2$ <statistical degrees of freedom>: the category counts have a fixed total, and one parameter was estimated. At the conventional $5\%$ level the <chi-squared distribution> with two <statistical degrees of freedom> has upper critical value $5.99$, so \b[the data do not reject the independent common-defect-probability model]. The approximate <p-value> is $e^{-X^2/2}\simeq0.146$; even at $10\%$, the critical value $4.61$ exceeds the observed statistic.
This is a goodness-of-fit conclusion rather than proof that individual bulbs are independent. The last expected count is only $4$, so the <chi-squared asymptotic approximation> is somewhat marginal in that cell. The stated calibration is the usual asymptotic test with all four cells retained; if more accurate small-count inference is required, calibrate the fitted statistic under the <binomial distribution> model. Pooling cells would require a corresponding treatment of parameter fitting and test calibration.
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