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Question: 6 balls are marked with numbers \(1\) to \(6\). If two balls are picked out of these \(6\) balls, wh...

6 balls are marked with numbers 11 to 66. If two balls are picked out of these 66 balls, what is the probability that the sum of the numbers on the balls is 88?
A. 1/15
B. 2/15
C. 1/5
D. 4/15
E. 1/3

Explanation

Solution

To find the required probability, we need to find the total number of outcomes and favorable number of outcomes first. Then, we will use the formula of probability that is Probability=Number of favorable outcomesTotal number of outcomes\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} and substitute the obtained value of favorable number of outcomes and total number of outcomes. After simplifying it, we will get the probability that the sum of the numbers on the ball is 88 when two balls are picked.

Complete step-by-step solution:
Since, the order is not important for selection of balls. So we will use the formula of combination to get the total number of outcomes as:
nCr=n!r!(nr)!\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\centerdot \left( n-r \right)!}
Where, nn is the total number of objects and rr is the number of objects chosen.
Now, we will substitute 66 for nn and 22 for rrin the above formula.
6C2=6!2!(62)!\Rightarrow {}^{6}{{C}_{2}}=\dfrac{6!}{2!\centerdot \left( 6-2 \right)!}
Solve the terms within the bracket.
6C2=6!2!(4)!\Rightarrow {}^{6}{{C}_{2}}=\dfrac{6!}{2!\centerdot \left( 4 \right)!}
We can write 15!15! as:
6C2=654!2!(4)!\Rightarrow {}^{6}{{C}_{2}}=\dfrac{6\centerdot 5\centerdot 4!}{2!\centerdot \left( 4 \right)!}
Here, we will cancel out the equal like terms and will expand the factorial terms as:
6C2=6512\Rightarrow {}^{6}{{C}_{2}}=\dfrac{6\centerdot 5}{1\centerdot 2}
Now, we will complete the multiplication in numerator and denominator as:
6C2=302\Rightarrow {}^{6}{{C}_{2}}=\dfrac{30}{2}
After simplifying the above step, we will have:
6C2=15\Rightarrow {}^{6}{{C}_{2}}=15
Since, there are only two combinations that sum is 88(2,6),(3,5). So, the number of favorable outcomes is 22.
Now, we will use the formula of probability to get the required probability as:
Probability=Number of favorable outcomesTotal number of outcomes\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}
Here, we will substitute the respective values as:
Probability=215\text{Probability}=\dfrac{\text{2}}{\text{15}}
Hence, 215\dfrac{\text{2}}{\text{15}} is the required probability that the sum of the numbers on the balls is88.

Note: Here a term is given as picking up the balls that means we have to select the balls and we use the combination for selection of objects. Combination is the possible number of outcomes of selecting objects where order doesn’t matter. The formula used for calculation of number of combination is:
nCr=n!r!(nr)!\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\centerdot \left( n-r \right)!}
Where, nrn\ge r and nn is the total number of objects and rr is the number of objects chosen.