# Mother, father and son line up at random for a family picture

E: son on one end,

F: Father in middle.

Determine P(E∣F).

Last updated date: 27th Mar 2023

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Answer

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Hint – In order to solve this problem we will first make a set of arrangements according to the condition provided in question then divide those elements in set E and F. Then we will solve P(E∣F) which is \[\dfrac{{{\text{P(E}} \cap {\text{F)}}}}{{{\text{P(F)}}}}\]. Doing this will take you to the right answer.

As we know the sign $ \cap $ indicates the common elements in the set E and F.

If mother (M), father (F) and son (S) line up for a family picture, then the sample space will be

S={MFS, MSF, FMS, FSM, SMF, SFM}

(Sample space is the set in which all the possible arrangement satisfying the condition occurs)

The set E in which the son is at one end can be:

$ \Rightarrow $ E = {MFS, FMS, SMF, SFM}

The set F in which the father is at the middle can be:

$ \Rightarrow $F = {MFS, SFM}

Number of elements common between E and F can be denoted as E∩F

Where,

E∩F = {MFS, SFM}

Probability of E∩F:

P(E∩F) = $\dfrac{2}{6}$=$\dfrac{1}{3}$

And the probability in which father is at middle is

P(F) = $\dfrac{2}{6}$ = $\dfrac{1}{3}$

As we know P(E∣F) = \[\dfrac{{{\text{P(E}} \cap {\text{F)}}}}{{{\text{P(F)}}}}\]

∴ P(E∣F) = \[\dfrac{{{\text{P(E}} \cap {\text{F)}}}}{{{\text{P(F)}}}}\]=$\dfrac{{\dfrac{1}{3}}}{{\dfrac{1}{3}}}$ = 1.

Hence the answer of this question is 1.

Note – Whenever you face such types of problems first you need to obtain the sample space of the arrangements and then divide the sample space with the different conditions. After that find the probability by using the formula number of favourable outcomes upon total number of outcomes and always remember that P(E∣F) = \[\dfrac{{{\text{P(E}} \cap {\text{F)}}}}{{{\text{P(F)}}}}\]where E∩F is the number of common elements in E and F.

__Complete step-by-step__solution__-__As we know the sign $ \cap $ indicates the common elements in the set E and F.

If mother (M), father (F) and son (S) line up for a family picture, then the sample space will be

S={MFS, MSF, FMS, FSM, SMF, SFM}

(Sample space is the set in which all the possible arrangement satisfying the condition occurs)

The set E in which the son is at one end can be:

$ \Rightarrow $ E = {MFS, FMS, SMF, SFM}

The set F in which the father is at the middle can be:

$ \Rightarrow $F = {MFS, SFM}

Number of elements common between E and F can be denoted as E∩F

Where,

E∩F = {MFS, SFM}

Probability of E∩F:

P(E∩F) = $\dfrac{2}{6}$=$\dfrac{1}{3}$

And the probability in which father is at middle is

P(F) = $\dfrac{2}{6}$ = $\dfrac{1}{3}$

As we know P(E∣F) = \[\dfrac{{{\text{P(E}} \cap {\text{F)}}}}{{{\text{P(F)}}}}\]

∴ P(E∣F) = \[\dfrac{{{\text{P(E}} \cap {\text{F)}}}}{{{\text{P(F)}}}}\]=$\dfrac{{\dfrac{1}{3}}}{{\dfrac{1}{3}}}$ = 1.

Hence the answer of this question is 1.

Note – Whenever you face such types of problems first you need to obtain the sample space of the arrangements and then divide the sample space with the different conditions. After that find the probability by using the formula number of favourable outcomes upon total number of outcomes and always remember that P(E∣F) = \[\dfrac{{{\text{P(E}} \cap {\text{F)}}}}{{{\text{P(F)}}}}\]where E∩F is the number of common elements in E and F.

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