A and B have less than 6 diamonds each.What is the probability

Question

A and B have less than 6 diamonds each.What is the probability that the sum of diamonds they have is more than 3 but less than 7?

A. 3/5
B. 9/25
C. 4/9
d. 2/3
E. 3/2

chetan2u · Answer

First let sfind the total ways.. 
 Since both A and B should have <6, they can have 0,1,2,3,4 or 5 so 6 ways each..
ways to choose one each = 6*6=36 ways.. 
 Here itself we can eliminate a,b and d as the denominator has to be 36 or its factor. d has 2 but prob>1, which is not possible

lets see how the sum can be 4, 5 or 6.. 
1) 4
 0,4 or 4,0 -2 ways
1,3 or 3,1 - 2 ways
2,2 --------- 1 way
Total - 5 ways

2) 5
 0,5 or 5,0 -2 ways
1,4 or 4,1 - 2 ways
2,3 or 3,2 - 2 way
Total - 6 ways

3) 6
 0,6 or 6,0 -2 ways -- NO, since we are looking for number<6 
 1,5 or 5,1 - 2 ways
2,4 or 4,2 - 2 way
3,3 -------- 1 way
Total - 5 ways

Total = 5+6+5 = 16 ways

so prob = \frac{16}{36} = \frac{4}{9} 
C

Bunuel · Answer

A has 0, 1, 2, 3, 4, or 5 diamonds;
B has 0, 1, 2, 3, 4, or 5 diamonds.

More than 3 but less than 7 = 4, 5, or 6.

4:
 {A, B} 
{4, 0}
{3, 1}
{2, 2}
{1, 3}
{0, 4}

5:
 {A, B} 
{5, 0}
{4, 1}
{3, 2}
{2, 3}
{1, 4}
{0, 5}

6:
 {A, B} 
{5, 1}
{4, 2}
{3, 3}
{2, 4}
{1, 5}

Favorable = 5 + 6 + 5 = 16.
Total = 6*6 = 36

Probability = 16/36 = 4/9.

Answer: C.

dhananjay1702 · Answer

Here should not the total number of possible ways in which A and B can have diamonds be 30? (0,0), (1,1)....(5,5) will be counted as one unlike (0,1) and (1,0).

Bunuel · Answer

They ARE counted as one. If you list all possibilities you'll be able to see that.

Kinshook · Answer

Given: A and B have less than 6 diamonds each.
Asked: What is the probability that the sum of diamonds they have is more than 3 but less than 7?

Number of diamonds A have = {0,1,2,3,4,5}: 6 cases
Number of diamonds B have = {0,1,2,3,4,5}: 6 cases

Total ways A & B can have diamonds = 6*6 = 36

The number of diamonds they have is more than 3 but less than 7 = {4,5,6}
Case 1: The sum of diamonds = 4
The number of ways such that the sum of diamonds they have is 4 = {(0,4),(1,3)(2,2),(3,1),(4,0)}: 5 cases
Case 2: The sum of diamonds = 5
The number of ways such that the sum of diamonds they have is 5 = {(0,5),(1,4)(2,3),(3,2),(4,1),(5,0)}: 6 cases
Case 3: The sum of diamonds = 6
The number of ways such that the sum of diamonds they have is 6 = {(1,5)(2,4),(3,3),(4,2),(5,1)}: 5 cases

The number of ways that the sum of diamonds they have is more than 3 but less than 7 = 5 + 6 + 5 = 16 cases

The probability that the sum of diamonds they have is more than 3 but less than 7 = 16/36 = 4/9

IMO C

Tchicaya48 · Answer

brunel  I have a question for you in the field of probability... Sometimes to find the number of possible outcomes we use a multiply and sometimes a addition I am confused please can u help me

Posted from my mobile device

Bunuel · Answer

OR probability: 
If Events A and B are independent, the probability that either Event A OR Event B occurs is:  P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B) .

This is basically the same as 2 overlapping sets formula:
 {total # of items in groups A or B} = {# of items in group A} + {# of items in group B} - {# of items in A and B} .

Note that if event are mutually exclusive then  P(A \ and \ B)=0  and the formula simplifies to:  P(A \ or \ B) = P(A) + P(B) .

Also note that when we say "A or B occurs" we include three possibilities:
A occurs and B does not occur;
B occurs and A does not occur;
Both A and B occur.

AND probability: 
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events:  P(A \ and \ B) = P(A)*P(B) .

This is basically the same as  Principle of Multiplication : if one event can occur in  m  ways and a second can occur independently of the first in  n  ways, then the two events can occur in  mn  ways.

22. Probability 
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Hope it helps.

A and B have less than 6 diamonds each.What is the probability

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A and B have less than 6 diamonds each.What is the probability

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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