(a) What is the probability of acquiring a amount of 7? (Enter youranswer together a fraction.)1(b) What is the probability of gaining a sum of 11? (Enter youranswer together a fraction.)2(c) What is the probability of gaining a sum of 7 or 11?(Enter your answer as a fraction.)Are these outcomes support exclusive?

Yes No

The concept of probability is provided to deal with this question.

You are watching: You roll two fair dice, one green and one red.

Probability is identified as an extent to which an event can occur. The probability of an event always lies between 0 and 1.

The probability will be calculated by generating the favorable outcomes and also sample room of one event.

Probability is defined as the number of outcomes produced from an occasion divided by the sample space. Sample space is characterized as the complete outcomes of one event.

P=NumberofoutcomesSamplespaceP = \frac\rmNumber of outcomes\rmSample spaceP=SamplespaceNumberofoutcomes​

Consider that 2 events, A and also B, take place with the probabilities P(A)andP(B)P\left( A \right)\rm and also P\left( B \right)P(A)andP(B) respectively. The probability the both the occasions occur is,

P(AandB)=P(A∩B)P\left( A\rm and also B \right) = P\left( A \cap B \right)P(AandB)=P(A∩B)

When both the occasions are independent,

P(AandB)=P(A∩B)=P(A)×P(B)\beginarrayc\\P\left( A\rm and B \right) = P\left( A \cap B \right)\\\\ = P\left( A \right) \times P\left( B \right)\\\endarrayP(AandB)=P(A∩B)=P(A)×P(B)​

The probability that either A or B wake up is,

P(AorB)=P(A∪B)=P(A)+P(B)−P(A∩B)\beginarrayc\\P\left( A\rm or B \right) = P\left( A \cup B \right)\\\\ = P\left( A \right) + P\left( B \right) - P\left( A \cap B \right)\\\endarrayP(AorB)=P(A∪B)=P(A)+P(B)−P(A∩B)​

The probability of an event constantly lies between 0 and 1.

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(a)

When two fair dice are rolled, one green and the various other red, climate the total number of outcomes (sample space) is 36.

The sample an are is together follows,

Samplespace=(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\rmSample space = \left\ \beginarrayl\\\left( \rm1,1 \right)\left( \rm1,2 \right)\left( \rm1,3 \right)\left( \rm1,4 \right)\left( \rm1,5 \right)\left( \rm1,6 \right)\\\\\left( \rm2,1 \right)\left( \rm2,2 \right)\left( \rm2,3 \right)\left( \rm2,4 \right)\left( \rm2,5 \right)\left( \rm2,6 \right)\\\\\left( \rm3,1 \right)\left( \rm3,2 \right)\left( \rm3,3 \right)\left( \rm3,4 \right)\left( \rm3,5 \right)\left( \rm3,6 \right)\\\\\left( \rm4,1 \right)\left( \rm4,2 \right)\left( \rm4,3 \right)\left( \rm4,4 \right)\left( \rm4,5 \right)\left( \rm4,6 \right)\\\\\left( \rm5,1 \right)\left( \rm5,2 \right)\left( \rm5,3 \right)\left( \rm5,4 \right)\left( \rm5,5 \right)\left( \rm5,6 \right)\\\\\left( \rm6,1 \right)\left( \rm6,2 \right)\left( \rm6,3 \right)\left( \rm6,4 \right)\left( \rm6,5 \right)\left( \rm6,6 \right)\\\endarray \right\Samplespace=⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)​⎭⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎫​

The probability of gaining the amount 11 is calculation as,

Numberofoutcomeshavingsumof7=(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)P(thesumoftwodicesare7)=NumberofoutcomesSamplespace=636=0.1666\beginarrayc\\\rmNumber the outcomes having sum of 7 = \left\ \left( \rm1,6 \right)\left( \rm2,5 \right)\left( \rm3,4 \right)\left( \rm4,3 \right)\left( \rm5,2 \right)\left( \rm6,1 \right) \right\\\\\P\left( \rmthe sum of 2 dices space 7 \right) = \frac\rmNumber of outcomes\rmSample space\\\\ = \frac\rm6\rm36\\\\ = \rm0\rm.1666\\\endarrayNumberofoutcomeshavingsumof7=(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)P(thesumoftwodicesare7)=SamplespaceNumberofoutcomes​=366​=0.1666​

(b)

The probability of getting sum of 11 once two dices are rolled is calculated as,

Numberofoutcomeshavingsumof11=(6,5)(5,6)P(thesumoftwodicesare11)=NumberofoutcomesSamplespace=236=0.055\beginarrayc\\\rmNumber that outcomes having actually sum of11 = \left\ \left( 6,5 \right)\left( 5,6 \right) \right\\\\\P\left( \rmthe sum of two dices room 11 \right) = \frac\rmNumber that outcomes\rmSample space\\\\ = \frac\rm236\\\\ = 0.055\\\endarrayNumberofoutcomeshavingsumof11=(6,5)(5,6)P(thesumoftwodicesare11)=SamplespaceNumberofoutcomes​=362​=0.055​

(c)

The probability of obtaining a sum of 7 or 11 is calculation as,

The probability of gaining the sum of 11 is calculate as,

Numberofoutcomeshavingsumof11=(6,5)(5,6)P(thesumoftwodicesare11)=NumberofoutcomesSamplespace=236=0.055\beginarrayc\\\rmNumber that outcomes having sum of11 = \left\ \left( 6,5 \right)\left( 5,6 \right) \right\\\\\P\left( \rmthe sum of 2 dices room 11 \right) = \frac\rmNumber of outcomes\rmSample space\\\\ = \frac\rm236\\\\ = \rm0\rm.055\\\endarrayNumberofoutcomeshavingsumof11=(6,5)(5,6)P(thesumoftwodicesare11)=SamplespaceNumberofoutcomes​=362​=0.055​

The probability of obtaining the amount of 7 is calculated as,

Numberofoutcomeshavingsumof7=(1,6)(6,1)(2,5)(5,2)(3,4)(4,3)P(thesumoftwodicesare7)=NumberofoutcomesSamplespace=636=0.1666\beginarrayc\\\rmNumber of outcomes having sum that 7 = \left\ \left( 1,6 \right)\left( 6,1 \right)\left( 2,5 \right)\left( 5,2 \right)\left( 3,4 \right)\left( 4,3 \right) \right\\\\\P\left( \rmthe sum of 2 dices are 7 \right) = \frac\rmNumberofoutcomes\rmSamplespace\\\\ = \frac636\\\\ = \rm0\rm.1666\\\endarrayNumberofoutcomeshavingsumof7=(1,6)(6,1)(2,5)(5,2)(3,4)(4,3)P(thesumoftwodicesare7)=SamplespaceNumberofoutcomes​=366​=0.1666​

The combined probability of gaining 7 or 11 is calculate as,

P(7or11)=P(getting7)+P(getting11)−P(getting7and11)=0.055+0.1666+0=0.2216\beginarrayc\\P\left( \rm7or11 \right) = P\left( \rmgetting 7 \right) + P\left( \rmgetting 11 \right) - P\left( \rmgetting 7 and 11 \right)\\\\ = \rm0\rm.055 + 0.1666 + 0\\\\ = \rm0\rm.2216\\\endarrayP(7or11)=P(getting7)+P(getting11)−P(getting7and11)=0.055+0.1666+0=0.2216​