Rashika is beginning a new job in human resources and expects to be

­Rashika is beginning a new job in human resources and expects to be assigned to several of the many human resources work groups. She has been told that there is a probability of ​​ that she will be assigned to the Recruitment Work Group and that there is a probability of ​​ that she will be assigned to the New Technology Work Group. She has no other information about the probabilities of the various possible assignments, or whether her assignment to one of these groups affects her chances of assignment to the other.

Select for Greatest probability for both assignments the greatest probability, compatible with the given probabilities, that Rashika will be assigned to both the Recruitment Work Group and the Technology Work Group. And select for Least probability for both assignments the least probability, compatible with the given probabilities, that Rashika will be assigned to both of these groups. Make only two selections, one in each column.­­­­

Calculating the greatest probability that she will be assigned to both groups is fairly straightforward.

The probability that she will be assigned to both groups can't be greater than lowest probability that she will be assigned to one of the groups, which is the probability that she will be assigned to the Recruitment Work Group, 1/2.

So, to calculate the greatest probability that she will be assigned to both groups, we can assume that, if she is assigned to the Recruitment Work Group, she is always assigned to the New Technology Work Group.

In that case, the probabllity that she will be assigned to both groups is 1/2.

For Greatest probability for both assignments, choose 1/2.

An intuitive way to calculate the least probability that she will be assigned to both is the following.

The maximum probability of any event occurring is 1. So, the maximum probability that she will be assigned to either the Recruitment Work Group or the New Technology Work Group is 1.

Now, since 1/2 + 3/4 > 1, there must be some times when both occur. In other words, the occurrences of the two events must overlap some of the time. After all, if they didn't overlap, the probability that one or the other would occur would be greater than 1.

If you think about it, the minimum overlap must be 1/4.

After all, if we fill up 1/2 of the possible total probability of 1 with the probability that she will be assigned to the Recruitment Work Group, then we have only 1/2 of the possible total probability of 1 left for the probability that she will be assigned to the New Technology Work Group. However, the probability that she will be assigned to the New Technology Work Group is 3/4. So, that 3/4 - 1/2 = 1/4 extra must overlap with the occurrence of her being assigned to the Recruitment Work Group.

In mathematical terms, P(X or Y) = P(X) + P(Y) - P(X and Y), and it must be the case that P(X or Y) ≤ 1.

So, if P(X) + P(Y) = 1/2 + 3/4 = 1 1/4, then to reduce 1 1/4 to the maximum P(X or Y) of 1, we need to subtract at least 1/4.

Thus, P(X and Y) must be at least 1/4, meaning the correct answer for Least probability for both assignments is 1/4.

For Least probability for both assignments, select 1/4.

Correct answer: 1/2, 1/4­­

Why can't least probability be 0?

 
­The only way for the probability to be 0 is for there to be a rule that, if Rashika is assigned to one of the two groups, she will for sure not be assigned to the other.

So, the probability can't be 0 because that rule cannot exist.

After all, if the probability is 1/2 that she will be assigned to the Recruitment Work Group, then for there to be a rule that she cannot be assigned to the New Technology Work group if she's assigned to the Recruitment Work Group, then the maximum probability that she will be assigned to the New Technology Work group would have to be 1/2.

In that case, we'd have a probability of 1/2 that she will be assigned to Recruitment and not New Technology and 1/2 that she will be assigned to New Technology and not Recruitment for a total probability of 1.

However, the probability that she will be assigned to the New Technology Work Group is not 1/2. It's 3/4. So, if the rule were that she cannot be assigned to the Recruitment Work Group if she's assigned to the New Technology Work Group, then there's no way the probability that she will be assigned to the Recruitment Work Group could be 1/2. It would have to be 1/4, since that's all that's left if we subtract 3/4 from the total probability of 1.

Thus, it's impossible for the probability to be less than 1/4.­

 
Probability behaves the same way as sets. We can maximize and minimize "Both" in exactly the same way.

P(Total) = P(A) + P(B) - P(Both) + P(Neither)

"Both" and "Neither" move together. When Both is max, Neither is also max. When Both is min, Neither is also min. 

The maximum probability of Both is simply the smaller of the two 1/2 and 3/4. If probability of allocation to Recruitment Work Group is 1/2, probability of allocation to both cannot exceed that. 
Select 1/2 in greater probability column. ANSWER

Both is minimum when Neither in minimum i.e. 0. Total must be 1.

1 = 1/2 + 3/4 - P(Both) + 0
P(Both) = 1/4

Select 1/4 in least probability column. ANSWER

Check this video for a detailed discussion on maximising and minimising in Sets: 
https://youtu.be/oLKbIyb1ZrI

 

­

Official Explanation

Infer
RO1
In terms of the probability x that Rashika will be assigned to both groups. For example, the probability that Rashika will be assigned to the Recruitment Work Group and not be assigned to the New Technology Work Group is ​open paren one half minus x close paren​, and the probability that Rashika will be assigned to the Recruitment Work Group is ​open paren one half minus x close paren plus x equals one half​. The two constraints that follow give all restrictions on the values that can appear in the regions in this graph: (1) Each value is between 0 and 1, inclusive. (2) The sum of the three values is between 0 and 1, inclusive. In particular, from (1) we have ​one half​ − x ≥ 0, or x ≤ ​one half​. Moreover, since x = ​one half​ satisfies both (1) and (2), it follows that the greatest possible value of x is ​one half​.

The correct answer is​one half​.

RO2
In the explanation above, from (2) we have ​open paren one half minus x close paren plus x plus open paren three fourths minus x close paren is less than or equal to 1​, or x ≥ ​one fourth​. Moreover, since x = ​one fourth​ satisfies both (1) and (2), it follows that the least possible value of x is ​one fourth​.

The correct answer is​one fourth​.

­                               Assigned to R        Not assigned to R          Total

Assigned to NT                  x                             a                       3/4

Not assigned to NT­                                                                   1/4

Total                               1/2                          1/2                      1


(i) Max
x <= \(\frac{1}{2}\) => Max x = \(\frac{1}{2}\)


(ii) Min
x min when a max

a <= \(\frac{1}{2}\) => Max a = \(\frac{1}{2}\)

=> Min x = \(\frac{3}{4} - \frac{1}{2} = \frac{1}{4}\)





 ­

Least = (50% + 75%) - 100% = 25%

Greatest = Min of (50%, 75%) = 50%­

Why are we assuming the neither to be 0 here? If we have some neither value, then the value of 1 will be further reduced that the min x will also be further reduced?
How do we know when to take neither as 0 or not incase of finding min x?
Thanks

P(a u b) = p(a) + p(b) - p(a n b)
1-neither = 1/2+3/4-both
Both = 1 1/4 - 1 + neither
Both = 1/4 + neither

both lowest when neither is lowest.

Would highly recommend anyone struggling with maximizing/minimizing problems to watch KarishmaB 's youtube explanation video(link below). I'm sure it'll make you much more comfortable with these qns. post watching this out.

https://youtu.be/oLKbIyb1ZrI

Happy to know it helped you!

I could not understand this as both events are mutually exclusive because it's mentioned that assignment in one group doesn't affect other , shouldn't the probability for both events happening be simple mutiplication 1/2 * 3/4 = 3/8. How the max min concept is applicable here

Same doubt why isn't the probability 1/2*3/4 for both happening. Please help Krunaal