The Gateway to Computer Science Excellence
0 votes
61 views
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) | 61 views
0
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

0 votes

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 (12.1k points)
0 votes
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)
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,645 questions
56,567 answers
195,748 comments
101,704 users