On her vacations Veena visits four cities (A,,B ,, C and D ) in a random order. probability that she

Hindi

14.Probability

medium

On her vacations Veena visits four cities $(A,\,B ,\, C$ and $D$ ) in a random order. What is the probability that she visits $A$ before $B$ and $B$ before $C ?$

$\frac{1}{6}$

Solution

The number of arrangements (orders) in which Veena can visit four cities $A,\,B,\,C$ or $D$ is $4 !$ i.e., $24 .$ Therefore, $n(S)=24$

since the number of elements in the sample space of the experiment is $24$ all of these outcomes are considered to be equally likely. A sample space for the experiment is

$S =\{ ABCD , \,ABDC , \,ACBD $, $ACDB , \,ADBC , \,ADCB$, $BACD,\, BADC,\, BDAC$, $BDCA, \,BCAD, ,BCDA,$ $CABD, \,CADB, \,CBDA$, $CBAD, \,CDAB, \,CDBA,$ $DABC,\, DACB,\, DBCA$, $DBAC, \,DCAB, \,DCBA\}$

Let the event 'Veena visits A before $B$ and $B$ before $C ^{*}$ be denoted by $F$.

Here $F =\{ ABCD , \,DABC , \,ABDC , \,ADBC \}$

Therefore, $P(F)=\frac{n(F)}{n(S)}=\frac{4}{24}=\frac{1}{6}$

Standard 11

Mathematics

For three non impossible events $A$, $B$ and $C$ $P\left( {A \cap B \cap C} \right) = 0,P\left( {A \cup B \cup C} \right) = \frac{3}{4},$ $P\left( {A \cap B} \right) = \frac{1}{3}$ and $P\left( C \right) = \frac{1}{6}$.

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normal

If $P(A) = 0.65,\,\,P(B) = 0.15,$ then $P(\bar A) + P(\bar B) = $

normal

A bag contains $3$ red and $5$ black balls and a second bag contains $6$ red and $4$ black balls. A ball is drawn from each bag. The probability that one is red and other is black, is

medium

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easy

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medium

On her vacations Veena visits four cities $(A,\,B ,\, C$ and $D$ ) in a random order. What is the probability that she visits $A$ before $B$ and $B$ before $C ?$

Solution

Similar Questions

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For three non impossible events $A$, $B$ and $C$ $P\left( {A \cap B \cap C} \right) = 0,P\left( {A \cup B \cup C} \right) = \frac{3}{4},$ $P\left( {A \cap B} \right) = \frac{1}{3}$ and $P\left( C \right) = \frac{1}{6}$.

The probability, exactly one of $A$ or $B$ occurs but $C$ doesn't occur is