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The probabilities that the independent events $A$, $B$, and $C$ will occur are: $\frac34$, $\frac12$, and $\frac14$.

What is the probability that at least one of the three events will happen?

My solution is

$1-$ Pr(intersection of the complements of $A$, $B$, $C$) thus $1-(\frac14\times\frac12\times\frac34)$

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  • $\begingroup$ Hi and welcome to the site! Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? $\endgroup$ – 5xum Sep 29 '14 at 12:27
  • $\begingroup$ What is the probability that none of the three events will happen? And what is the complement of that event? $\endgroup$ – drhab Sep 29 '14 at 12:28
  • $\begingroup$ you might want to respond to the questions asked in the comments above, else it is possible that your questions is closed.. $\endgroup$ – Seyhmus Güngören Sep 29 '14 at 12:33
  • $\begingroup$ I have already edited my question wwith my sollution. I hope this will help. $\endgroup$ – Jonarie Ramos Vergara Sep 29 '14 at 12:37
  • $\begingroup$ Your solution is okay. $\endgroup$ – drhab Sep 29 '14 at 12:49
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Hint: That probability is the same with $p*=1-p(\mbox{ none of the events will happen})$

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