16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T469,{"@context":"https://schema.org","@type":"Quiz","name":"Mht Cet 2019 Probability PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/mht-cet/questions/2019/mathematics/probability/m2txy6q5","educationalLevel":"Entrance exam preparation","about":"Mht Cet / Mathematics / Probability","hasPart":{"@type":"Question","name":"Mht Cet 2019 Probability PYQ with Solution","text":"In a bionomial distribution, mean is 18 and variance is 12 then p= ...........","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"B"},"comment":{"@type":"Comment","text":"In a binomial distribution, the mean (\\mu) and variance (\\sigma^2) are given by the formulas: \\mu = np \\sigma^2 = np(1-p) where n is the number of trials and p is the probability of success in each trial. Given: Mean, \\mu = 18 Variance, \\sigma^2 = 12 Substituting the mean into the formula, we have: 18 = np For the variance, substitute and rearrange: 12 = np(1-p) Now, we solve these two equations simultaneously. From the mean: n = \\frac{18}{p} Substitute n from Equation 1 into Equation 2: 12 = \\..."}}}18:T429,

In a binomial distribution, the mean ($\mu$) and variance ($\sigma^2$) are given by the formulas:

$$ \mu = np $$

$$ \sigma^2 = np(1-p) $$

where $ n $ is the number of trials and $ p $ is the probability of success in each trial.

Given:

Mean, $\mu = 18$

Variance, $\sigma^2 = 12$

Substituting the mean into the formula, we have:

$$ 18 =

Probability

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