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Let $A$ and $B$ be events, $P(A) = \frac{1}{4} $, $P(A\cup B) = \frac{1}{3} $ and $ P (B) = p $.

  1. Find $p$, if $A$ and $B$ are mutually exclusive.
  2. Find $p$, if $A$ and $B$ are independent.
  3. Find $p$, if $A$ is a subset $B$.

Can someone help me to solve it?

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  • $\begingroup$ Though there are hints now, what have you tried? Please tag the home work questions as such; don't worry--you'll get well thought out hints that should help you complete the solution. Regards, $\endgroup$ – user21436 May 4 '12 at 11:53
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    $\begingroup$ It's curious how much this reminds me of this problem where several people agreed that the solution was "not mathematical". $\endgroup$ – MJD May 4 '12 at 12:59
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Hint.1: $P(A \cup B) = P(A) + P (B)$ when the two events are mutually exclusive.

Hint.2: $P(A \cap B) = P(A) \cdot P (B)$ when the two events are independent.

Hint.3: $P(A \cap B) = P(A) $ when $A$ is a subset of $B$.

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  • $\begingroup$ So, ok, I find this result in I) 1 = $P(A) + P(B)$, SO ... 1 = $\frac{1}{4} + $p SO ... p = $\frac{1}{1} - \frac{1}{4} = \frac{3}{4}$ $\endgroup$ – mastergoo May 4 '12 at 13:01
  • $\begingroup$ That is right ? $\endgroup$ – mastergoo May 4 '12 at 13:07
  • $\begingroup$ That's right @mastergoo. Well done. $\endgroup$ – Gigili May 4 '12 at 13:17
  • $\begingroup$ II) $1 = P(A) * P(B)$ ... ... $1 = \frac{1}{3} * p$ ... ... $p = \frac{1}{\frac{1}{3}}$ ... ... $p = 3$ $\endgroup$ – mastergoo May 4 '12 at 13:46
  • $\begingroup$ II) That's right ? $\endgroup$ – mastergoo May 4 '12 at 13:48
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  • $ A$ and $B$ are mutually exclusive: $P(A \cap B)= 0$
    $P(A \cup B) = P(A) + P(B)$
    Then you can find value of $p$ using it.

  • $A$ and $B$ are independent: $P(A \cap B)= P(A) \cdot P(B)$
    $P(A \cup B) = P(A) + P(B)-P(A) \cdot P(B)$
    Then you can find value of p using it.

  • $A$ is a subset of $B$, that is, $A \subseteq B$: $A \cup B = B$
    That implies $P(A \cup B) = P(B)$
    Then you can find value of p using it.

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  • $\begingroup$ Dear Prasad, +1; Happy to see you contributing. I wanted to let you know that the site allows TeX mark up for the Math. Please TeXify your posts to add to your well-thought out answers. Regards, $\endgroup$ – user21436 May 4 '12 at 11:50
  • $\begingroup$ Thank you for your suggestion and i am learning $\endgroup$ – Prasad G May 4 '12 at 11:54
  • $\begingroup$ How can I solve III) ? $\endgroup$ – mastergoo May 4 '12 at 14:05
  • $\begingroup$ Can you help-me ? $\endgroup$ – mastergoo May 4 '12 at 17:35
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If $A$ and $B$ are independent events then $p(A \cap B) = p(A) \cdot p(B)$ and note that

  • $p(A \cup B) = p(A) + p(B) - p(A \cap B)$

This answers ii

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  • $\begingroup$ How can I solve III) ? $\endgroup$ – mastergoo May 4 '12 at 14:05
  • $\begingroup$ @mastergoo: See the first answer $\endgroup$ – user9413 May 4 '12 at 14:05
  • $\begingroup$ Ok, I saw but I can not understand. Logicaly if A is subset of B, A U B = B. $\endgroup$ – mastergoo May 4 '12 at 14:07
  • $\begingroup$ @mastergoo: what does $A \subset B$ mean, it means all elements in $A$ are in $B$. Now, from this you have$A \cap B = A$, and hence $p(A\cap B) = p(A)$ $\endgroup$ – user9413 May 4 '12 at 14:09
  • $\begingroup$ Oh, OK, I understand now.. My solution for I and II are right ? $\endgroup$ – mastergoo May 4 '12 at 14:11

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