Strategy guide 4 , page 195 , chapter 12 , 4th edition

[krishnakumar.gopinathan]

Hi,

This problem is from MGMAT study guide 4 (4th edition) pg 195.

Kevin has wired 6 light bulbs to a board so that, when he presses a button each bulb has an equal chance of lighting up or staying dark. Each of the six bulbs is independent of the other five.

a. In how many different configurations could the bulbs on the board light up (including the configuration in which none of them light up) ?

b. In how many configurations could exactly three of the bulbs light up ?

c. When Kevin presses the button what is the probability that exactly three of the bulbs light up ?

My doubt is in the explanation given for part a.
The answer given is that since each bulb has two options on or off the total number of options is 2^6 = 64.

why can't i consider taking
(no of ways of exactly 1 bulb lighting up) *
(no of ways of exactly 2 bulbs lighting up) *
(no of ways of exactly 3 bulbs lighting up) *
....
(no of ways of exactly 6 bulbs lighting up)

which is 6C1 * 6C2 * 6C3....

I am confused here..


Thanks in advance.

Krishna Kumar

[mschwrtz]

Two mistakes, one strategic and one substantive.

The strategic mistake is that there is no reason to think of this as a very elaborate n choose r problem. Simply use the Fundamental Counting Principal, which tells us that we may determine the number of ways to combine choice/outcomes/whatever by multiply the number of options at each decision point.

Hey, the language in our Strategy Guide might be clearer, Fundamental Counting Principle: If you must make a number of separate decisions, then multiply the number of ways to make each individual decision to find the number of ways to make all the decisions.

Since you have six different decision here, each with two options, there are 2^6 ways to combine decisions.

OK, but that just shows that your method is inefficient, not that it's wrong. Here's the substantive problem. You neglected 6 choose 0. If you toss that in, you'll have 1+6+15+20+15+6+1 or 64 configurations.

[krishnakumar.gopinathan]

Hi,

Thanks a lot.

I realize that I did make another mistake too. I used 'and' instead of 'or'.

krishna kumar

[jnelson0612]

Thanks for your clarification krishna. I'm glad the problem now makes more sense to you.

Thank you,

[lionelpq]

Another question related to this problem: In part c, since the probability of the light bulb turning on is .5, why wont the probability of 3 lighting up is 1/2? (i.e 6*.5)?

Thanks

[jnelson0612]

Another question related to this problem: In part c, since the probability of the light bulb turning on is .5, why wont the probability of 3 lighting up is 1/2? (i.e 6*.5)?

Thanks

Hi Lionel,
Think about it this way:

1) First, how many possible arrangements of bulbs on and off do I have? The first bulb can be either on or off (2 possibilities), the second bulb can be either on or off (2 possibilities), and so on. So I have 2*2*2*2*2*2 or 2^6, which is 64 possible arrangements of bulbs on and off.

2) How many of those will have three bulbs on? Let's call an on bulb "Y" and and off bulb "N". Thus, how many ways can I arrange the letters YYYNNN? Using the anagram method, I have 6!/(3!3!). This equals 20 ways I can arrange three Ys and three Ns.

3) I can conclude that I have a 20/64 probability of have three on bulbs, or 5/16.