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if P(E1E2'E3)=1/8, P(E1E2E3')=1/12, P(E1'E2'E3')=1/12 where E1,E2,E3 are independent events. Find 1)P(E1) 2)P(E1'E2E3')

probability-theory

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asked yesterday

Rupali ShekhawatRupali Shekhawat

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put on hold as off-topic by Kavi Rama Murthy, John Omielan, José Carlos Santos, RRL, Saad yesterday

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-2

1

$begingroup$

if P(E1E2'E3)=1/8, P(E1E2E3')=1/12, P(E1'E2'E3')=1/12 where E1,E2,E3 are independent events. Find 1)P(E1) 2)P(E1'E2E3')

probability-theory

share|cite|improve this question

asked yesterday

Rupali ShekhawatRupali Shekhawat

1

New contributor

Rupali Shekhawat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by Kavi Rama Murthy, John Omielan, José Carlos Santos, RRL, Saad yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, John Omielan, José Carlos Santos, RRL, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried so far?
  $endgroup$
  – Mindlack
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

if P(E1E2'E3)=1/8, P(E1E2E3')=1/12, P(E1'E2'E3')=1/12 where E1,E2,E3 are independent events. Find 1)P(E1) 2)P(E1'E2E3')

probability-theory

share|cite|improve this question

asked yesterday

Rupali ShekhawatRupali Shekhawat

1

New contributor

Rupali Shekhawat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked yesterday

Rupali ShekhawatRupali Shekhawat

1

New contributor

Rupali Shekhawat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked yesterday

Rupali ShekhawatRupali Shekhawat

1

New contributor

Rupali Shekhawat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Rupali ShekhawatRupali Shekhawat

1

New contributor

Rupali Shekhawat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Rupali ShekhawatRupali Shekhawat

1