When two six-sided die arc rolled, the probability of getting a one on both is 1

Question

When two six-sided die arc rolled, the probability of getting a one on both is 1/36. This means that a) of every 36 rolls, exactly 1 will have both die be one. b) the odds against both die being one arc 36 to 1. c) in the long run, the average number of ones is 1/36. d) in the long run the outcome that both die arc one will occur on 1/36 of all rolls. An experiment has four possible outcomes: A, B, C, and D. Which of the following is a legitimate assignment of probabilities for these four events? a) 0.2, 0.2, 0.2, 0.2 b) 0.3, 0.3, 0.3, 0.1 c) 0.6, 0.2, 0.3, -0.1 d) 0.4, 0.4, 0.4, 0.4 Choose an American household at random and ask how many televisions that household owns. Here are the probabilities as determined by a recent survey: Number of Televisions 0 12 3 Probability 0.10 0.44 0.28 0.18 What is the probability that a randomly chosen household owns at least two televisions? a) 0.18 b) 0.28 c) 0.46 d) 0.54 e) 0.82

Explanation / Answer

1. Answer : b

Probability definition = Number of favorable events /Total number of events

Herer the favorable events for both dice show 1 is : 1

And Total number of events : 36

So the odds against both die being 1 are 36 to 1

The odds for both die being 1 are 1 to 36

2. An experiment has 4 possibe outcomes A,B,C,D. Which of the following is the legitimate statement of probabilities of these four events.

Right answer . b

As A,B,C,D are the four outcomes of an experiment then when that experiment is conducted either A or B or C or D has to occur.

Total event space S = A,B,C,D

P(S) = P(A or B or C or D) = P(A) + P(B) +P(C) +(D) =1

only (b) satisfy that condition as P(A) + P(B) +P(C) +(D) =0.3+0.3+0.3+0.1 = 1

3.Probability that randomly choosen house owns atleast Two televisions.

Right Answer . C

Probability that randomly choosen house owns atleast Two televisions

= Probability that randomly choosen house owns Two televisions or Probability that randomly choosen house owns three televisions

=Probability that randomly choosen house owns Two televisions +  Probability that randomly choosen house owns three televisions

From the table

Probability that randomly choosen house owns Two televisions = 0.28

Probability that randomly choosen house owns three televisions = 0.18

Probability that randomly choosen house owns atleast Two televisions

  =Probability that randomly choosen house owns Two televisions +  Probability that randomly choosen house owns three televisions

= 0.28+0.18 = 0.46

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