Problem Solving

A single particle is accelerated through a magnetic field which causes it to take on either a clockwise or counterclockwise spin. If the particle takes on a clockwise spin it is called left-handed. If it takes on a counterclockwise spin, it is called right-handed. An experiment calls for 5 particles to be accelerated through the field, one at a time. If the probability of a particle taking on a clockwise spin is 2/5 , and the probability of a particle taking on a counterclockwise spin is 3/5, then what is the probability that more particles will be right-handed than left-handed?

varun289 wrote:A single particle is accelerated through a magnetic field which causes it to take on either a clockwise or counterclockwise spin. If the particle takes on a clockwise spin it is called left-handed. If it takes on a counterclockwise spin, it is called right-handed. An experiment calls for 5 particles to be accelerated through the field, one at a time. If the probability of a particle taking on a clockwise spin is 2/5 , and the probability of a particle taking on a counterclockwise spin is 3/5, then what is the probability that more particles will be right-handed than left-handed?

A. 38(3/5)^3
B. (3/5)^3(79/25)
C. (3/5)^4 (2/5)
D. (3/5)^3 (2/5)^2
E. (3/5)^3 (19/25)

Let R = right-handed and L = left-handed.
P(R) = P(counter-clockwise) = 3/5.
P(L) = P(clockwise) = 2/5.

For the outcome to be favorable, the number of R's must be greater than the number of L's.

Case 1: 3 R's, 2 L's.
P(RRRLL) = 3/5 * 3/5 * 3/5 * 2/5 * 2/5 = (3/5)³ * (2/5)².
RRRLL is only ONE WAY to get 3 R's and 2 L's.
To account for ALL OF THE WAYS, we must multiply by the number of ways to arrange the letters RRRLL (5!/3!2!):
5!/(3!2!) * (3/5)³ * (2/5)² * = 10 * (3/5)³ * (2/5)²

Case 2: 4 R's, 1 L
P(RRRRL) = 3/5 * 3/5 * 3/5 * 3/5 * 2/5 = (3/5)^4 * (2/5).
RRRRL is only ONE WAY to get 4 R's and 1 L.
To account for ALL OF THE WAYS, we must multiply by the number of ways to arrange the letters RRRRL (5!/4!):
5!/4! * (3/5)^4 * (2/5) = 5 * (3/5)^4 * (2/5)

Case 3: 5 R's
P(RRRRR) = 3/5 * 3/5 * 3/5 * 3/5 * 3/5 = (3/5)^5

Since anyone of these 3 cases -- Case 1 OR Case 2 OR Case 3 -- would yield a favorable outcome, we ADD the results above:
[10 * (3/5)³ * (2/5)²] + [5 * (3/5)^4 * (2/5)] + (3/5)^5

= (3/5)³ [(10)(2/5)² + (5)(3/5)(2/5) + (3/5)²]

= (3/5)³ (40/25 + 30/25 + 9/25)

= (3/5)³(79/25).

The correct answer is B.

until i looked at answer choices i thought this was another one of those trick questions about probability not changing.

just so i have a differnetial to go by how would you word this problem to be like that??

GMATGuruNY wrote:

varun289 wrote:A single particle is accelerated through a magnetic field which causes it to take on either a clockwise or counterclockwise spin. If the particle takes on a clockwise spin it is called left-handed. If it takes on a counterclockwise spin, it is called right-handed. An experiment calls for 5 particles to be accelerated through the field, one at a time. If the probability of a particle taking on a clockwise spin is 2/5 , and the probability of a particle taking on a counterclockwise spin is 3/5, then what is the probability that more particles will be right-handed than left-handed?

A. 38(3/5)^3
B. (3/5)^3(79/25)
C. (3/5)^4 (2/5)
D. (3/5)^3 (2/5)^2
E. (3/5)^3 (19/25)

Let R = right-handed and L = left-handed.
P(R) = P(counter-clockwise) = 3/5.
P(L) = P(clockwise) = 2/5.

For the outcome to be favorable, the number of R's must be greater than the number of L's.

Case 1: 3 R's, 2 L's.
P(RRRLL) = 3/5 * 3/5 * 3/5 * 2/5 * 2/5 = (3/5)³ * (2/5)².
RRRLL is only ONE WAY to get 3 R's and 2 L's.
To account for ALL OF THE WAYS, we must multiply by the number of ways to arrange the letters RRRLL (5!/3!2!):
5!/(3!2!) * (3/5)³ * (2/5)² * = 10 * (3/5)³ * (2/5)²

Case 2: 4 R's, 1 L
P(RRRRL) = 3/5 * 3/5 * 3/5 * 3/5 * 2/5 = (3/5)^4 * (2/5).
RRRRL is only ONE WAY to get 4 R's and 1 L.
To account for ALL OF THE WAYS, we must multiply by the number of ways to arrange the letters RRRRL (5!/4!):
5!/4! * (3/5)^4 * (2/5) = 5 * (3/5)^4 * (2/5)

Case 3: 5 R's
P(RRRRR) = 3/5 * 3/5 * 3/5 * 3/5 * 3/5 = (3/5)^5

Since anyone of these 3 cases -- Case 1 OR Case 2 OR Case 3 -- would yield a favorable outcome, we ADD the results above:
[10 * (3/5)³ * (2/5)²] + [5 * (3/5)^4 * (2/5)] + (3/5)^5

= (3/5)³ [(10)(2/5)² + (5)(3/5)(2/5) + (3/5)²]

= (3/5)³ (40/25 + 30/25 + 9/25)

= (3/5)³(79/25).

The correct answer is B.

After going through various such questions on this website, I could solve this one with ease. But how does one solve such questions in under 2 mins? Hard questions are like a double edged sword- they fetch you good scores if you answer them correctly, but they also consume more time (which could have adverse repercussions to the score).

Perfect 700-800 Question.

You're not expected to solve this and if you do try to solve it you'll be there all day and probably get it wrong. There is a higher probability of a particle being right handed than left so the probability of more right than left is greater than 1/2. Only one answer is between 1/2 and 1. B, done, move on. This is not a math test and if you treat it like one you will perform poorly. The harder the question the greater the probability that 4 of your answers are impossible, leaving one.

Hi GMATGuru,

I am a bit confused here.

if we are already testing 5 particles and the probability that right handed ones are 3/5 and the probability that left handed ones is 2/5, we already know for a fact that we have 3 right handed particles and 2 left handed particles.

Hence why did we assume that we could have 4 right handed particles and 1 left handed, and also 5 right handed and 0 left handed.

i thought if we know something for a fact then the probability is one, and since we have 3 right handed particles and 2 left handed particles, then the probability that right handed particles are more than left handed particles should be one.

can you please explain?

thanks

GMATGuruNY wrote:

varun289 wrote:A single particle is accelerated through a magnetic field which causes it to take on either a clockwise or counterclockwise spin. If the particle takes on a clockwise spin it is called left-handed. If it takes on a counterclockwise spin, it is called right-handed. An experiment calls for 5 particles to be accelerated through the field, one at a time. If the probability of a particle taking on a clockwise spin is 2/5 , and the probability of a particle taking on a counterclockwise spin is 3/5, then what is the probability that more particles will be right-handed than left-handed?

A. 38(3/5)^3
B. (3/5)^3(79/25)
C. (3/5)^4 (2/5)
D. (3/5)^3 (2/5)^2
E. (3/5)^3 (19/25)

Let R = right-handed and L = left-handed.
P(R) = P(counter-clockwise) = 3/5.
P(L) = P(clockwise) = 2/5.

For the outcome to be favorable, the number of R's must be greater than the number of L's.

Case 1: 3 R's, 2 L's.
P(RRRLL) = 3/5 * 3/5 * 3/5 * 2/5 * 2/5 = (3/5)³ * (2/5)².
RRRLL is only ONE WAY to get 3 R's and 2 L's.
To account for ALL OF THE WAYS, we must multiply by the number of ways to arrange the letters RRRLL (5!/3!2!):
5!/(3!2!) * (3/5)³ * (2/5)² * = 10 * (3/5)³ * (2/5)²

Case 2: 4 R's, 1 L
P(RRRRL) = 3/5 * 3/5 * 3/5 * 3/5 * 2/5 = (3/5)^4 * (2/5).
RRRRL is only ONE WAY to get 4 R's and 1 L.
To account for ALL OF THE WAYS, we must multiply by the number of ways to arrange the letters RRRRL (5!/4!):
5!/4! * (3/5)^4 * (2/5) = 5 * (3/5)^4 * (2/5)

Case 3: 5 R's
P(RRRRR) = 3/5 * 3/5 * 3/5 * 3/5 * 3/5 = (3/5)^5

Since anyone of these 3 cases -- Case 1 OR Case 2 OR Case 3 -- would yield a favorable outcome, we ADD the results above:
[10 * (3/5)³ * (2/5)²] + [5 * (3/5)^4 * (2/5)] + (3/5)^5

= (3/5)³ [(10)(2/5)² + (5)(3/5)(2/5) + (3/5)²]

= (3/5)³ (40/25 + 30/25 + 9/25)

= (3/5)³(79/25).

The correct answer is B.

Amrabdelnaby wrote:Hi GMATGuru,

I am a bit confused here.

if we are already testing 5 particles and the probability that right handed ones are 3/5 and the probability that left handed ones is 2/5, we already know for a fact that we have 3 right handed particles and 2 left handed particles.

Hence why did we assume that we could have 4 right handed particles and 1 left handed, and also 5 right handed and 0 left handed.

i thought if we know something for a fact then the probability is one, and since we have 3 right handed particles and 2 left handed particles, then the probability that right handed particles are more than left handed particles should be one.

can you please explain?

thanks

According to the prompt:
That probability of a particle taking on a clockwise spin (and thus being left-handed) is 2/5, while the probability of a particle taking on a counterclockwise spin (and thus being right-handed) is 3/5.
Implication:

The probability that the first particle is left-handed = 2/5.
The probability that the first particle is right-handed = 3/5.

The probability that the second particle is left-handed = 2/5.
The probability that the second particle is right-handed = 3/5.

The probability that the third particle is left-handed = 2/5.
The probability that the third particle is right-handed = 3/5.

The probability that the fourth particle is left-handed = 2/5.
The probability that the fourth particle is right-handed = 3/5.

The probability that the fifth particle is left-handed = 2/5.
The probability that the fifth particle is right-handed = 3/5.

It is possible that all of the particles are left-handed.
It is possible that all of the particles are right-handed.
It is possible that some of the particles are left-handed, while the rest are right-handed.
We do not know for a fact that, of the 5 accelerated particles, 2 will be left handed and 3 will be right-handed.