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A man alternately tosses a coin and throws a dice, beginning with the coin. Then probability that he will get a head before he gets a 5 or 6 on dice is

1) 1/4

2) 3/4

3) 4/5

4) 4/7
in Probability by Active (4.9k points) | 59 views
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is it B ?

H+H'P'(AUB)H+H'P'(AUB)H'P'(AUB)H+..... Here PROB OF GETTING 5 OR 6 ON DICE = 1/6+1/6 =1/3 AND NOT OF THIS LET P'(AUB)= 1-1/3=2/3

1/2+1/2*2/3*1/2+1/2*2/3*1/2*2/3*1/2+....

= 3/4

2 Answers

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since the man alternatively throws a coin and than tossess a coin where coin is tossed first..

so the sequence becomes = H,(5/6) + T,(1,2,3,4),H,(5,6) +  T,(1,2,3,4) ,T,(1,2,3,4) ,H,(5,6)..................

                                         =1/2 * 2/6 + 1/2*4/6 *1/2*2/6+................

                                        =1/6 + 1/3 *1/6 +1/3*1/3*1/6+.......

                                        =1/6( 1/(1-1/3) )= 1/4 ANSWER

by Boss (12k points)
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Correct answer is 3/4.

Probability =( prob. Of getting heads) or (prob. Of getting tail and prob. Of not getting 5,6 and then prob. Of getting heads) or (prob. Of getting tails and prob. Of not getting 5,6 and prob. Of getting tails and prob. Of not getting 5,6 and prob. Of getting heads) or ......

I think you get the idea.

So, this would be a sum of geometric progression.

Probability = 1/2 + 1/2 * 4/6 * 1/2 + 1/2 * 4/6 * 1/2 * 4/6 * 1/2 + ........

Notice how "and" & "or" in the story expression becomes * & + respectively in the mathematical expression.

Now the sum of Geometric progression with r<1 is :

Probability = 1/2 * (1/(1 - 1/2*4/6)) = 1/2 * 3/2 = 3/4.
by (339 points)

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