probability

Question

Let x be the number obtained from rolling a fair dice and you toss an unbiassed coin X times. What is the probablity that X=5 given that you have obtained 3 heads from X tosses?

Comments

5/16

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Ans)  5/16

Answer 1

My Answer is 2/7.Given that 3 heads appeared it means the outcome of die roll is at least 3.Now if the out come of die roll is 3, then ways of occurring 3 heads is 3C3, if 4 occurred in die roll then ways of occurring 3 heads is 4C3, similarly 5C3 and 6C3.so all possible ways in which 3 heads can appear is (3C3 + 4C3 + 5C3 + 6C3). The favourable outcome is when die roll happens to be a 5 and 3 heads appear that is 5C3. So desired probability will be (5C3)/(3C3 + 4C3 + 5C3 + 6C3) which gives 2/7.

Comments

What is the ans Pooja? I think 2/7 is crrct?

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@anurag answer will be 5/16

you need to consider probability of heads and tails also i.e for 3 heads in 5 tosses

5C3 (1/2)3 (1/2)2 =5C3 (1/2)5

Answer 2

Ans is 5/16.

The probability of getting 3 heads when the dice outcome is 5 is 1/6*10/32.

Probability of getting 3 heads when the dice outcome is x is: for x:3, P(3head|x=3)=1/8 , x:4, P(3head|x=4)=4/16, x:5,

P(3head|x=5)=10/32, x:6, P(3head|x=6)=20/64.

So the Probability is (1/6*10/32)/(1/8 + 4/16 + 10/32 + 20/64)=5/16. 

 

Answer 3

Here probabili of head=1/2 and probability of no head=1/2.

so by binomial distribution of probability we get the probability= 5 C 3(1/2)^3*(1/2)^2=5/16