فی دائما ثابت F(1) = P(Y = 1) = 3 احتمالات . = (2 = PY = (2) علمود احل نفس القبلة 90 W 12 21 4 نيستخدم من القانون 0 Ex: PE PS == 6 2 21 y 0 1 2 f(y) = P(Y=y) 1/7 4/7 2/7 القانون نحل بطريقة وبعدها اعوض بهذا القانون) PE n! تحل بطريقة بسيطة () = X! (n-1) Example (4) Έχαι رب Solution: Ply=s) A shipment of 20 similar laptop computers to a retail outlet contains 3 (2)Find that are defective. If a school makes a random purchase of 2 of these computers, find the probability distribution for the number of defectives. Let X be a random variable whose values x are the possible numbers of defective computers purchased by the school. Then x can only take the numbers 0, 1, and 2. ناطر PO P IF P 136 Solu 200 (0) = P(X = 0) f(2) = P(X=2) (29) W f(1) = P(X = 1) = 51 190 (29) 190 الجدول معلم fl 0 1 2 (29) 190 f(x) = P(X=x) 136/190 51/190 3/190 م را تاكد السبط اقل فى المقام Explain this question in detail.

فی دائما ثابت F(1) = P(Y = 1) = 3 احتمالات . = (2 = PY = (2) علمود احل نفس القبلة 90 W 12 21 4 نيستخدم من القانون 0 Ex: PE PS == 6 2 21 y 0 1 2 f(y) = P(Y=y) 1/7 4/7 2/7 القانون نحل بطريقة وبعدها اعوض بهذا القانون) PE n! تحل بطريقة بسيطة () = X! (n-1) Example (4) Έχαι رب Solution: Ply=s) A shipment of 20 similar laptop computers to a retail outlet contains 3 (2)Find that are defective. If a school makes a random purchase of 2 of these computers, find the probability distribution for the number of defectives. Let X be a random variable whose values x are the possible numbers of defective computers purchased by the school. Then x can only take the numbers 0, 1, and 2. ناطر PO P IF P 136 Solu 200 (0) = P(X = 0) f(2) = P(X=2) (29) W f(1) = P(X = 1) = 51 190 (29) 190 الجدول معلم fl 0 1 2 (29) 190 f(x) = P(X=x) 136/190 51/190 3/190 م را تاكد السبط اقل فى المقام Explain this question in detail.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter1: Fundamental Concepts Of Algebra

Section1.2: Exponents And Radicals

Problem 41E

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