#probability_doubt_in_basic_concept

Question

1) Two coins are tossed.What is the probability of getting one head and one tail ? WKT ANS IS 2/4=1/2

2) A coin is tossed twice.What is the probability of getting one head and one tail ? WKT ANS is 2/4=1/2

3) A bag has 3 red balls and 4 green balls.Two balls are picked from the bag randomly.What is the probability of getting one red and one green ? ANS GIVEN IS (3/7*4/6)....WHY NOT (3/7*4/6)+(4/7*3/6)

I MEAN WHY WE ARE CONSIDERING THE ORDER FOR 1) AND 2) BUT NOT FOR 3)

THAT IS IN 1) AND 2) ......{HEAD,TAIL} AND {TAIL,HEAD} ARE TREATED DIFFERENTLY BUT IN 3) BOTH {R,G} AND {G,R} ARE TREATED SAME......WHY ???

please explain I am getting confused here ....

Comments

The relationship between picking a ball and getting head or tail accordingly should be mentioned which is I think not mentioned in the question.

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I think you are getting confused between selection and arrangement! The game here is getting 2 different balls so either you can write 4/7*3/6 or 3/7*4/6 but not together, if you write it together it may become ordered selection i.e. first red and then green or first green and then red!

Hope, you get it!

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@habib can u see now bro ...i have edited the 3 rd question ....

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@smartmeet In all 3 questions, I dont think we are speaking about arrangements ...please give me more clarity ...i am not getting your explanation ...

Answer 1

Second QS:

• We have 4 events
• All the events are equally likely with probability $ p = 0.25$
• Therefore total probability = $\sum p = 1$
• Required probability = $\begin{align*} &\frac{p+p}{\sum p} = 0.5 \\ \end{align*}$

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First QS:

• Almost same as Second QS:
• Required probability = $\begin{align*} &\frac{p+p}{\sum p} = 0.5 \\ \end{align*}$

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Third QS:

• $4$ possible events.
• They are not all equally likely.
• $\begin{align*} &p_1 = \frac{3}{7}*\frac{2}{6} \;,\; p_2 = \frac{3}{7}*\frac{4}{6} \;,\; p_3 = \frac{4}{7}*\frac{3}{6} \;,\; p_4 = \frac{4}{7}*\frac{3}{6} \\ \end{align*}$
• $\begin{align*} \sum p_i = 1 \end{align*}$
• Required probability  = $\begin{align*} \frac{p_2+p_3}{\sum p_i} = \frac{4}{7} \end{align*}$

Comments

Wow !! How do you do this ?

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free time..use as many colours as possible :)