Mr.William bought a box of 15 tube lights out of which exactly 5 were

Question

Mr.William bought a box of 15 tube lights out of which exactly 5 were broken. He took out 4 tube lights at random. What is the probability that exactly 2 out of 4 tube lights be selected were broken?

A. 3/91
B. 13/91
C. 31/91
D. 1/13
E. 30/91

KanishkM · Accepted Answer

Total Tube lights = 15
Broken = 5
Not Broken = 10

Exactly 2 out of 4 tube lights be selected were broken
=  \frac{{5C2 * 10C2}}{15C4}

= 30/91

UB001 · Answer

Given :
Total Tubelights = 15
No. of broken Tubelights = 5 Exactly 
No. of unbroken Tubelights = 10
No. of Tubelights taken out = 4

Question: 
Probability of Exactly 2 out of 4 Tubelights taken out were broken =
(Probability of 2 Broken Tubelights AND Probability of 2 Unbroken Tubelights ) x Its Arrangement

= (5/15).(4/14).(10/13).(9/12).4! / (2!.2!)
= 30/91
Answer Option E is Correct..!

Posted from my mobile device

Kinshook · Answer

Given: Mr.William bought a box of 15 tube lights out of which exactly 5 were broken. He took out 4 tube lights at random. 
Asked: What is the probability that exactly 2 out of 4 tube lights be selected were broken?

Total ways = 15C4 = 15*7*13 = 15*91
Favorable ways = 10C2*5C2 = 45*10 = 450

Probability = 15*30/15*91 = 30/91

IMO E

Abir77 · Answer

Great explaination

Abir77 · Answer

Great explanation

MathRevolution · Answer

Total: 15
Broken: 5
Not Broken: 10

Chosen at random: 4

Total outcomes:   ^{15}\mathrm{C_4}  = 1,365

Desired Outcome:  Exactly 2 out of 4 tube lights be selected were broken: Select 2 from 5 (broken) and the remaining 2 from not broken(10).

=>  ^{5}\mathrm{C_2}  *  ^{10}\mathrm{C_2}

=> 10 * 45

=> 450

Probability :  \frac{Desired outcome }{ Total outcome}

=>  \frac{450 }{ 1365 }

=>  \frac{30}{91}

Answer E

rbmgengenharia · Answer

To find the probability that exactly 2 out of the 4 tube lights selected are broken, we can use the binomial probability formula. This formula is given by:

P(X=k) = (n choose k) * p^k * (1-p)^(n-k)

where:

P(X=k) is the probability of getting k successes in n trials
n is the total number of trials (in this case, the number of tube lights selected)
k is the number of successes (in this case, the number of broken tube lights selected)
p is the probability of success in a single trial (in this case, the probability of selecting a broken tube light)
In this problem, n = 4 (since 4 tube lights were selected), k = 2 (since 2 of the tube lights selected were broken), and p = 5/15 (since 5 of the 15 tube lights were broken). Plugging these values into the formula, we get:

P(X=2) = (4 choose 2) * (5/15)^2 * (10/15)^2 = 6 * (1/3)^2 * (2/3)^2 = (1/9) * (4/9) = 4/81

Therefore, the probability that exactly 2 out of the 4 tube lights selected are broken is 4/81.

ajaysureshkumar · Answer

How is 4! / (2!.2!) arrived? I understood the former part of (5/15).(4/14).(10/13).(9/12)

NikhilNideeshwar · Answer

In this method why haven't we considered the different ways of choosing broken and working light bulbs. why is it only selection? Aren't we supposed do 4!/2!x2! x 5c2 x 10c2 ? THANKS 😊

NikhilNideeshwar · Answer

hey ajaysureshkumar, we are asked to find p( exactly 2 out of 4 tube lights broken)....so draw 4 dashes - broken , broken , working , working (this method is the same way we solve word arrangement problems).

to find total no. of arrangements we do 4!/2! x 2! ( we do this to avoid any internal permutations from taking place). I think you can carry on after this part.

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Mr.William bought a box of 15 tube lights out of which exactly 5 were

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