geeks for geeks

Question

In a bunch of 13 T-shirts only one is of Medium size, which is correct fit for the searching person. Each time wrong size is picked, the person throws it away and pick the next T-shirt. What is the probability that the correct size T-shirt can be searched in 8th attempt ?

Comments

1/13

Answer 1

you have 13 T-shirts in which 12 are not matching and 1 is matching.

 

at first attempt, you have to take non-matching t-shirt, probability = $\frac{12}{13}$

in second attempt, you have to take non-matching t-shirt, probability = $\frac{11}{12}$ ( due to you have 12 t-shirts only available with you )

in third attempt, you have to take non-matching t-shirt, probability = $\frac{10}{11}$ ( due to you have 11 t-shirts only available with you )

in fourth attempt, you have to take non-matching t-shirt, probability = $\frac{9}{10}$ ( due to you have 10 t-shirts only available with you )

in fifth attempt, you have to take non-matching t-shirt, probability = $\frac{8}{9}$ ( due to you have 9 t-shirts only available with you )

in sixth attempt, you have to take non-matching t-shirt, probability = $\frac{7}{8}$ ( due to you have 8 t-shirts only available with you )

in seventh attempt, you have to take non-matching t-shirt, probability = $\frac{6}{7}$ ( due to you have 7 t-shirts only available with you )

in eighth attempt, you have to take matching t-shirt, probability = $\frac{1}{6}$ ( due to you have 6 t-shirts only available with you )

Therefore total probability = $ \frac{12 * 11 * 10 * 9 * 8 * 7 * 6 * 1}{ 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 }$ = $ \frac{1}{ 13 }$

Comments

but answer is

Let L_i be the event of searching medium sized T-shirt in i^th attempt. P(L'₁ ∩ L'₂ ∩ L'₃ ∩ L'₄ ∩ L'₅ ∩ L'₆ ∩ L'₇ ∩ L₈ ) = P(L'₁) . P(L'₂ / L'₁) . P(L'₃ / L'₁ ∩ L'₂) ------- P( L₈ / L'₁ ∩ L'₂ ...... L'₇) = 8/13 . 7/12. 6/11. 5/10. 4/9. 3/8. 2/7. 1/6 = 40320 / 51891840 = 1 / 1287

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Their Formula is correct but values are not correct...

P(L'1 ) should be $\frac{12}{13}$ ... They did wrong. consider my answer

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can you please explain the answer ,how this formula is used???

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Basic Point :-

P(A/B) = $\frac{P(A∩B)}{B} $, where P(A/B) means Probability of A when B is Happened.

 

Note that in the question they given that Each time wrong size is picked, the person throws it away

it means probability of sucess or unsucess is changed every time,

How that formula derived ?

Select first one which is Non-Matched ===> P(L'₁ )

seclect second one which is non-Matched but given that you already selected L'₁ . ===>  P(L'₂/L'₁ )

seclect third one which is non-Matched but given that you already selected L'₁ and L'₂. ===>  P(L'₃/ ( L'₁ ∩ L'₂ ) )

 

and so on....

 

Substitute this formula, P(A/B) = $\frac{P(A∩B)}{B} $ ===> you can get the answer

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great explaination @ Shaik Masthan 

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thank you @ Shaik Mastan