Higher Algebra  1
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Question 1 of 5
1. Question
One bag contains 5 white balls and 3 red balls, and a second bag contains 4 white and 5 red balls. From one of them, chosen at random, two balls are drawn: what is the probability that they are of different colors?
Correct
P(event) = P(Bag 1) P(Different Balls) + P(Bag 2) P(Different Balls)
=1/2×5_{C1}3_{C1}8_{C2 }+ 1/2 ×5_{C1}4_{C1}9_{C2}
=1/2((15/28)+5/9)
=275/504Incorrect
P(event) = P(Bag 1) P(Different Balls) + P(Bag 2) P(Different Balls)
=1/2×5_{C1}3_{C1}8_{C2 }+ 1/2 ×5_{C1}4_{C1}9_{C2}
=1/2((15/28)+5/9)
=275/504 
Question 2 of 5
2. Question
Laura has to arrange 4 different balls in 3 different boxes and she doesn’t care for their order. Please help her find the difference in the number of ways in which she can execute her task, if empty boxes are allowed and not allowed.
Correct
This question is based on different to different (D to D) distribution
No. of ways, if empty boxes are allowed = 3^{4} = 81
No. of ways, if empty boxes are not allowed = 3^{4} – 3c_{1}.2^{4} + 3c_{2}. 1^{4} = 36
Difference = 8136 = 45.Incorrect
This question is based on different to different (D to D) distribution
No. of ways, if empty boxes are allowed = 3^{4} = 81
No. of ways, if empty boxes are not allowed = 3^{4} – 3c_{1}.2^{4} + 3c_{2}. 1^{4} = 36
Difference = 8136 = 45. 
Question 3 of 5
3. Question
Three dice are rolled simultaneously. If the sum of the numbers that appear is not less than 6, what is the probability that the sum is equal to 16
Correct
The table below shows the number of ways in which different sums can be obtained when 3 dice are rolled simultaneously .
Incorrect
The table below shows the number of ways in which different sums can be obtained when 3 dice are rolled simultaneously .

Question 4 of 5
4. Question
In an election for the post of college representatives, every student can vote in 56 ways. Every student votes for at least one candidate. If the total number of candidates exceeds the maximum number that can be elected by 2, then find the number of candidate.
Correct
Let the number of candidates be n so that the number of college representatives to be elected is (n – 2). Now as man can vote for 1, 2, 3…. or (n – 2) candidates.
Hence number of ways in which one can vote is ^{n}C_{1} +^{ n}C_{2} + ^{n}C_{3} + ……. + ^{n}C_{(n – 2)} = 56
Add and subtract^{ n}C_{0} + ^{n}C_{(n – 1)} + ^{n}C_{n} in the above equation
2^{n}– (1 + ^{n}C_{1} + ^{n}C_{0}) = 56
2^{n}– (1 + n + 1) = 56
2^{n}– n = 56
By inspection, n = 6 satisfy the above equation.Incorrect
Let the number of candidates be n so that the number of college representatives to be elected is (n – 2). Now as man can vote for 1, 2, 3…. or (n – 2) candidates.
Hence number of ways in which one can vote is ^{n}C_{1} +^{ n}C_{2} + ^{n}C_{3} + ……. + ^{n}C_{(n – 2)} = 56
Add and subtract^{ n}C_{0} + ^{n}C_{(n – 1)} + ^{n}C_{n} in the above equation
2^{n}– (1 + ^{n}C_{1} + ^{n}C_{0}) = 56
2^{n}– (1 + n + 1) = 56
2^{n}– n = 56
By inspection, n = 6 satisfy the above equation. 
Question 5 of 5
5. Question
A die is thrown. If it shows a six, we draw a ball from a bag consisting 2 black balls and white six balls and keep it out of bag. If it does not shown six, then we toss a coin. Then, the probability of getting a black and a head consecutively is
Correct
P(Event) = P(Black)P(Head)
P(Black) = P(6)P(Black Ball) = 1/6×[C(2,1)/C(8,1)] =1/24
P(Head) = P(Not 6) P(Head) = (5/6)×(1/2)=5/12
P(Event) = (1/24)×(5/12)=5/288Incorrect
P(Event) = P(Black)P(Head)
P(Black) = P(6)P(Black Ball) = 1/6×[C(2,1)/C(8,1)] =1/24
P(Head) = P(Not 6) P(Head) = (5/6)×(1/2)=5/12
P(Event) = (1/24)×(5/12)=5/288