# Five horses are in a race. Mr. A selects two of the horses at random and bets on  them. The probability that Mr. A selected the winning horse is_________________.

Aman Bansal
592 Points
9 years ago

Dear Shiva,

The probability of winning is 2/5

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums.

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

upender surepally
126 Points
9 years ago

2/5

Shaurya Gupta
41 Points
9 years ago

total no of options - 5

selected options - 2

therefore probability that Mr.A selected the winning horse - 2/5

Aravind Bommera
36 Points
9 years ago

The probability of winning is 2/5

ishaan agarwal
8 Points
8 years ago
This question came in aieee 2003. 2/5 is not in the option only.
R Surya Narayan
13 Points
5 years ago
In total there are 5 horses. out of these Mr A selects two horses. This can be done in $\binom{5}{2}$(5C2)=10 ways.
Out of the five horses only one horse wins. Let us say Mr A has selected the horse. He has to select one more horse so that his set of two has the winning horse. out of remaining 4 one can be selected in $\binom{4}{1}$ways.Thus
$n(E)=\binom{4}{1}=4$
$n(S)=\binom{5}{2}=10$
$\Rightarrow P(E)=\frac{n(E)}{n(S)}=\frac{4}{10}=\frac{2}{5}$

ankit singh
2 years ago
In total there are 5 horses. out of these Mr A selects two horses. This can be done in 5C2=10 ways.
Out of the five horses, only one horse wins.
Let us say Mr A has selected the horse. He has to select one more horse so that his set of two has the winning horse. out of remaining 4 one can be selected in 4C1=4 ways.
Thus,

In total there are 5 horses. out of these Mr A selects two horses. This can be done in $\binom{5}{2}$(5C2)=10 ways.
Out of the five horses only one horse wins. Let us say Mr A has selected the horse. He has to select one more horse so that his set of two has the winning horse. out of remaining 4 one can be selected in $\binom{4}{1}$ways.Thus
$n(E)=\binom{4}{1}=4$
$n(S)=\binom{5}{2}=10$
$\Rightarrow P(E)=\frac{n(E)}{n(S)}=\frac{4}{10}=\frac{2}{5}$

The probability that Mr.A selected the winning horse is = 104=52In total there are 5 horses. out of these Mr A selects two horses. This can be done in 5C2=10 ways.
Out of the five horses, only one horse wins.
Let us say Mr A has selected the horse. He has to select one more horse so that his set of two has the winning horse. out of remaining 4 one can be selected in 4C1=4 ways.
Thus,
The probability that Mr.A selected the winning horse is = 104=52
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}