Probability

Question

A fair coin is tossed 3 times in succession.If the first toss produced a head then the probability of getting exactly two heads in 3 tosses(including the first toss ) is

• A. 1/8
• B. 3/8
• C. 1/2
• D. 3/4

Comments

First toss is head ... now we have to get exactly one head in rest two toss .. 

1. 2nd is head and 3rd is tail. ( probability = (1/2)*(1/2) )

2. 2nd is tail and 3rd is head. ( probability = (1/2)*(1/2) )

so resultant probability = 1/2.

Ans- C

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this is conditional probability based problem bcz question is saying if one event happened already then what will be the probability of happen second event  which sample space is based on  earlier event...

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You mean - probability of getting exactly one head in 2nd and 3rd toss when head in 1st toss is already occurred.

Let probability of getting exactly one head in 2nd and 3rd toss = p(A) = 1/2 (HT, TH)

 Probability of getting head in 1st toss = p(B) = 1/2.

So resultant prob. = p(B) * p(A) / (  p(B) * p(A) +  p(B) * p(A') )

                              = (1/2 ) * (1/2) / ( (1/2) * (1/2) + (1/2) * (1/2) )

                              = (1/4) / (1/2)

                              = 1/2. 

 

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we can think something like initially total sample space={ HHH ,HHT,HTH,HTT,THH,THT,TTH,TTT} now favourable  outcomes for event when first toss produced a head={HHH,HTH,HHT,HTT}  ...

.again we have to find the probability for getting exactly two head when already first toss produced a head so for this case sample space= {HHH,HTH,HHT,HTT}  and favourable outcome ={HHT,HTH} so probability of getting this will be 2/4=1/2

Answer 1

here are only 4 outcomes in the sample space 
 

HHH HHT HTH HTT,

 

So, Probability of getting exactly two heads =$\frac{2}{4}=\frac{1}{2}$ 

 

Hence,Option(c)$\frac{1}{2}$ is the correct choice.

 

Answer 2

$P(A/B) = \frac{P(A \cap B)}{P(B)}$

Where $P(A/B)$ = Probability of A after B has happened.

$P(A \cap B)$ = probability of happening A and B together.

$P(A)$ = Probability of happening event A.( Event : In a toss no, of heads is = 2 )

$P(B)$ = Probability of happening event B. ( Event : First toss is head )

In this questions, $P(A \cap B)$ = 1/4 ( HTH and HHT out of  8 possible cases)

and $P(B)$ = 1/2 ( HTT,HTH,HHT,HHH out of all 8 possible cases )

Here we can count manually since no. of tosses are less. For large case we need to use combination formule to make counting simple

Comments

got it

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I was doing mistake Ignoring the word. "exactly"

Answer 3

DIFFERENT WAY OF SOLVING:

Actually We can easily Visualise this things with Diagramatic Approach

This is how we can do in one another way

Answer 4

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