Question

A father purchased dress for his 3 daughters. The dresses are of same color but different size and they are kept in dark room. What is probability that all the 3 will not choose their own dress?
(A) \[2/3\]
(B) \[1/3\]
(C) \[1/6\]
(D) \[1/9\]

Answer 1

Hint: Probability - Probability is numerical description of how likely an event is to occur.
The probability of an event is a number between 0 and 1, where, 0 indicates impossibility of the event and 1 indicates certainty.

For example when we flip a coin: There are two possible outcomes—heads or tails.
50% chances of getting heads and 50$\% $ of getting tails.
But also we can find out this by using probability formula.
Probability$ = \dfrac{{favourable\;outcomes}}{{total\,outcomes}}$
Here in our case favourable outcome for heads as well as tails is 1.
(As a coin can have only one heads and only one tails)
Total number of outcomes $ = $ no. of heads + no of tails
So, total outcomes \[ = 1 + 1 = 2\]

Now, we will find probability of getting heads on coin flip.
For that we will use above formula and from formula we have
Probability $ = \dfrac{{favourable\;outcomes}}{{total\,outcomes}}$
Therefore, $P = \dfrac{1}{2}$
Which is \[50\% \] chances for both.
Lets solve above question.

Complete step by step answer:
In above question, 3 girls are selecting one-one dress for them which are not their own dress.
So,
(A) When 1st girl came she has three choices and she will choose a wrong dress.
Here, total outcomes = total number of dresses \[ = \] 3
Favorable outcomes $ = $ number of wrong
Dresses $ = $ 2
Because 1 dress is for her which is the right dress for her so rest 2 dresses are wrong.
So,
Probability, P1 of that girl to choose wrong dress out of 3 is
$ = \dfrac{{favourable\;outcomes}}{{total\,outcomes}}$
$ = \dfrac{2}{3}$
Now,
(B) 2nd girl came she has two choices. She choose wrong dress and out of two only one dress is for her and one is wrong.
Total outcomes $ = $ total number of dresses $ = $ 2
Favorable outcomes $ = $ number of wrong
Dresses $ = $ 1
So probability, P2 of that girl to choose wrong dress out of 2 is,
$ = \dfrac{{favourable\;outcomes}}{{total\,outcomes}}$
$ = \dfrac{1}{2}$
(C) For 3rd girl only one dress is left and which is wrong.
Therefore, probability, P3 of choosing wrong dress $ = $ 1
So, Probability,P that all the 3 will not choose their own dress is,
\[P = P1 \times P2 \times P3\]
$ = \dfrac{2}{3} \times \dfrac{1}{2} \times 1$
$ = \dfrac{2}{6} = \dfrac{1}{3}$
Therefore, Probability that all the 3 will not choose their own dress $ = \dfrac{1}{3}$
Option (B) i.e. \[1/3\] is correct option

Note: Probability ranges between 0 and 1.

First we need to make a set(S) for total number of outcomes before solving the problem.
Like
¡. When a coin is tossed, \[S = \left\{ {H,T} \right\}\]
¡¡. When two coins are tossed, \[S = \left\{ {HH,HT,TH,TT} \right\}\]
where H $ = $ Head and T $ = $ Tail