WebIf P(A)=0.4,P(B)=p,P(A∪B)=0.6 and A and B are given to be independent events , find the value of p . Medium Solution Verified by Toppr P(A∪B)=P(A)+P(B)−P(A∩B) ⇒0.6=0.4+p−P(A∩B) ⇒P(A∩B)=0.4+p−0.6=p−0.2 Since , A and B are independent events. ∴P(A∩B)=P(A)×P(B) ⇒p−0.2=0.4×p ⇒p−0.4p=0.2 ⇒0.6p=0.2 ⇒p= 0.60.2= 31 Was this … Webif `P (A)=0.4,P (B^ (\'))=0.6` and `P (AcapB)=0.15` then the value of `P ( Doubtnut 2.7M subscribers Subscribe 2 401 views 3 years ago if `P (A)=0.4,P (B^ (\'))=0.6` and `P...
P-Value And Statistical Significance: What It Is & Why It Matters
WebIf P (A and B) =0.5, P (A) = 0.7, and P (B) =0.6, find P (A or B). Homework: 11.5 OR and A Score: 0 of 1 pt 11.5.15 Find the indicated probability. If P (A or B)=0.8, P (A)=0.5, and P … Webwhy create a profile on Shaalaa.com? 1. Inform you about time table of exam. 2. Inform you about new question papers. 3. New video tutorials information. just beachy designs
Probability Calculator Mathway
WebProbability of a Normal Distribution. Use the calculator below to find the area P shown in the normal distribution, as well as the confidence intervals for a range of confidence levels. … Web25 mrt. 2024 · The null hypothesis (H0): μ = 2 ounces. The alternative hypothesis: (HA): μ ≠ 2 ounces. The auditor conducts a hypothesis test for the mean and ends up with a p … Web12 sep. 2024 · 0.6 is the answer; this is evident if you draw partially overlapping circles A & B. Place probability 0.3 in the common area (A∩B) and 0.4 in the rest of circle A. Place probability 0.1 in the part of B that excludes the A∩B area. That leaves 0.2 as outside all circles; you have four different probabilities that add up to 1.0. Upvote • 0 Downvote just beachy burgers