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In a lottery , a girl chooses seven different natural numbers at random from 1 to 22 and if these seven numbers match with the seven numbers already fixed by the lottery committee , she wins the prize. what is the probability of winning the prize in the game ?

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Native English Speaker with 32 years experience in teaching English and communication skills.

There are 22C7 ways. (7^22)C= (20×19×18×17×16×15×14)/(1×2×3×4×5×6×7)= 77520 So, Probability of winning the lottery = 1 / 77520
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SME in Mathematics & Statistics

Total number of choices = 22C7 = 170544. Out of this only 1 will match with the prefixed number. Hence the required probability is 1/170544.
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

There are 22 natural numbers, 7 numbers may be chosen in 22C7 ways.. (7^22)C= (20×19×18×17×16×15×14)/(1×2×3×4×5×6×7)= 77520 there is only 1 favourable case which is being fixed by lottery committee. Probability of winning the lottery = 1 / 77520
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Physics and Maths made easy

Dear Akshay, Sorry for the slip of hand mistake in my previous answer. The right answer is (1/170544) as,22 C 7= (22!)/ = 170544. Sorry for the inconvenience.
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Excel Tutor

1/22c7=1/170544
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

There are 22 natural numbers, 7 numbers may be chosen in 22C7 ways.. (7^22)C= (22X21X20×19×18×17×16)/(1×2×3×4×5×6×7)= 170544 Since, there is only 1 favourable case which is being fixed by lottery committee. therefore,Probability of winning the lottery = 1 / 170544
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Tutor

There are total 22C7choices that she has to choose from out of which only one will win her the prize. So, the chance of winning a prize is 1/(22C7) = 1/170544
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Bachelor of Engineering (B.E.) in Electronics & Communication

1 / 77520
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it is 1/22C7
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Tutor

n(s)=22c7 required prob=1/22c7
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