PROBABILITY

Q1. A dice is thrown once, what is the probability of 3 showing up?

Ans1. Total outcomes = 6, favorable outcomes = 1

            Probability = 1/6

Q2. Two dice are thrown, which event is more probable a score of 11 or 4?

Ans2. When two dice are thrown total outcomes are 36

For 11, favorable outcomes = 2 (6, 5) (5, 6),

Therefore probability = 2/36 = 1/18

For 11, favorable outcomes = 2 (6, 5) (5, 6),

Therefore probability = 2/36 = 1/18

Q3. Two dice are thrown, what is the probability of getting two 6’s?

Ans3. Here favorable outcome is one (6,6)

Total outcomes = 36

Probability = 1/36

Q4. Two coins are tossed. What is the probability of having 2 heads?

Ans4. The Total outcomes are 4 (HH , HT , TH , TT)

The favorable outcomes is 1 (HH)

Probability = ¼

Q5. A card is selected from a pack of 52 cards. Find the probability that it is an ace or a spade?

Ans5. There are 13 spades which includes one ace, and another 3 aces in the deck

            Therefore, favorable outcomes = 16

            Total Outcomes = 52

            Probability = 16/52 = 4/13

Q6. A card is selected from a pack of 52 cards .find the probability that it is a spade or a ace or a king.

Ans6. Number of spades = 13, number of aces = 4, number of kings = 4

            Now spades have one king and one ace, so

            Number of aces left = 3, Number of kings left = 3

            Probability = 13/52 + 3/52 + 3/52 = 19/52

Q7. A bag contains 2 violet, 3 black and 4 green balls. Find the probability that a ball drawn at random will be violet or green.

Ans 7.   Total balls are 9, with violet = 2, Black = 3 and Green = 4

            Probability of violet ball = 2/9

            Probability of green ball = 4/9

            P (violet or green) = 6/9

Q8. A bag contains 3 violet, 3 black and 3 green balls. If three balls are taken out with replacement. Find the probability of:

1.      All are green

2.      None is green

3.      All are of same colour

Ans8. Total balls are 10, with violet = 3, Black = 3 and Green = 3

The experiment is happening with replacement, which means each time a ball is taken out it is being replaced, so total balls and respective number of balls stay the same

            Probability of violet ball = 3/9 = 1/3

            Probability of black ball = 3/9 = 1/3

            Probability of green ball = 3/9 = 1/3

1.      All are green

P(all green) =  1/3 x 1/3 x 1/3 = 1/27

2.      None is green

Now P(not green) = 6/9 = 2/3

And P(none are green) = 2/3 x 2/3 x 2/3 = 8/27

3.      All are same colour

All are same colour means = 3 violets or 3 blacks or 3 greens

            = (1/3 x 1/3 x 1/3) + (1/3 x 1/3 x 1/3) + (1/3 x 1/3 x 1/3)

            = 3/27 = 1/9

Q9. A bag contains 3 violet, 3 black and 3 green balls. If three balls are taken out without replacement. Find the probability of:

1.      All are green

2.      None is green

3.      All are of same colour

Ans9. Total balls are 10, with violet = 3, Black = 3 and Green = 3

The experiment is happening without replacement, which means each time a ball is taken out it is not being replaced, so total balls and respective number of balls are reducing as balls are being taken out

            Probability of violet ball = 3/9

            Probability of black ball = 3/9

            Probability of green ball = 3/9

1.      All are green

P(all green) =  3/9 x 2/8 x 1/7 = 6/504

2.      None is green

Now P(not green) = 6/9

And P(none are green) = 6/9 x 5/8 x 4/7 = 120/504

3.      All are same colour

All are same colour means = 3 violets or 3 blacks or 3 greens

            = (3/9 x 2/8 x 1/7) + (3/9 x 2/8 x 1/7) + (3/9 x 2/8 x 1/7)

            = 18/504

Q10. Ravi is going for a blood test; his chance of being positive for a disease is 0.1, to confirm again and again, he takes three tests. What is the probability that

1.      He will be positive in all the tests

2.      He is be positive at least once

Ans10.  Chance of being positive = 0.1

            Chance of being positive in three tests = 0.1 x 0.1 x 0.1 = 0.001

            Chance of being negative = 1 – 0.1 = 0.9

             Chance of being negative in three tests = 0.9 x 0.9 x 0.9 = 0.729

            Chance of being positive at least once = 1 – Chance being negative in all tests

                                                                          = 1 – 0.729 = 0.271

Q11. There are 6 blue marbles and 4 red marbles. What is the probability of your drawing a blue marble and then my drawing a red one?

Ans11. The language is everything in a question; student should be able to understand that this is a question of conditional probability. Here, what is chance of drawing a red ball, when a blue ball has already been drawn?

Which is P(R/B), which is P(R and B)/P(R)

Now P(R and B) = 6/10 x 4/9 = 24/90 = 4/15

And P(R) = 6/10, therefore P(R/B) = (4/15)/(6/10) =  4/9

Q12. Assume that a test to detect a disease whose prevalence is (1/1000) has a false positive rate of 5% and a true positive rate of 100%. What is the probability that a person found to have a positive result actually has the disease assuming that you know nothing about the person’s symptoms?

Ans12. H = has the disease, P = Test result is positive

As prevalence is 1/1000, therefore P(H) = .001, therefore P(Not H) = .999

As false rate is 5%, P(P/not H) = 0.05

And true positive is 100%, therefore P(T/H) = 1.00

Now we have to find if person has the disease in case he has tested positive already, which is P(H/P)

P(H/P) = P(H and P)/P(P)

Now P(H and P) = 0.001 x 1 = 0.001

P(P) =  P(P and H) + P(P and not H)

        = 0.001 + (.999)x (.05)

        = .05095

P (H/P) = .001/.05095 = .019627

Q13. A box contains 6 white balls and 4 black balls and another box contains 4 white balls and 6 black balls. Find the probability that a ball selected from one of the box is a white ball.

Ans 13. Now, In Box I, 6 white and 4 black balls are there

            Probability of white ball from box I = 6/10

            Probability of selecting box I = ½

            P(white ball from box I) = ½ x 6/10 = 6/20

Now, In Box II, 4 white and 6 black balls are there

            Probability of white ball from box I = 4/10

            Probability of selecting box I = ½

            P(white ball from box II) = ½ x 4/10 = 4/20

Therefore, P(white ball from box I or box II) = 4/20 + 6/20 =1/2

Q14. A box contains 6 white balls and 4 black balls and another box contains 4 white balls and 6 black balls. Find the probability of selecting a white ball from box I.

Ans14. Students should be able to understand the difference in this question and last question, in this question the chance of drawing a white ball from Box I has been asked, which is clear case of conditional probability.

Now, In Box I, 6 white and 4 black balls are there

            Probability of white ball from box I = 6/10

            Probability of selecting box I = ½

            P(white ball from box I) = ½ x 6/10 = 6/20

Now, In Box II, 4 white and 6 black balls are there

            Probability of white ball from box I = 4/10

            Probability of selecting box I = ½

            P(white ball from box II) = ½ x 4/10 = 4/20

Therefore, P(white ball from box I or box II) = 4/20 + 6/20 =1/2

Here P(Box I/white ball) = P(Box I and white ball) / P(white ball)

                                         = (6/20) / (1/2) = 6/40 = 3/20

Q15. From a deck of 52 cards, one card is lost; the next two cards drawn are spades, what is the probability that the lost card was a spade?

Ans15. There are 52 cards, In case the card lost was a spade, then its probability at that time would be 13/52, the other cards to be spades will be 12/51 and 11/50. In case that card is not spade, the probability of other cards being spade is 13/52 and 12/51 (students may think how this is possible). From here the probability of third card to be spade is 11/50

Alternatively, assume you got the third card lying some where around, you added the card to the deck without seeing which card it is, now what is the probability of card being spade, again 11/50.

Q16.   In GMAT verbal section there are 41 questions. Each question has 5 options out of which only one is correct. If someone clicks answers at random, what is the probability that he will get 20 out of the 41 questions correct ?

Ans16. P(x) = nCx × px × q(n-x)

          41C20 × (1/5)20 × (4/5)21

  Q17.The probability that a graduate student being male is 0.25 and that being female is 0.75. The probability that a male student passes the course is 0.7 and that a female student does it is 0.80. A student selected at random is found to have completed the course. What is the probability that the student is (i) male and (ii) female?

Ans17. Probability of being a male = 0.25

      Probability of being a Female = 0.75

      Probability for female to complete the course = 0.80

Probability for male to complete the course = 0.70

Probability for a male who has done the course = 0.25 × 0.7 = 0.175

Probability for a female who has done the course = 0.75 × 0.8 = 0.6

Probability for a person who has done the course is either a male finishing the course or a female finishing the course = 0.25 × 0.7 + 0.75 × 0.8 = 0.775

Probability a student being male who has finished the course = 0.175/0.775 = 0.225

Probability a student being Female who has finished the course = 0.6/0.775 = 0.774

Q18. A, B and C in order cut a pack of cards replacing them after each cut, on condition that first who cuts a spade shall win a prize. find their respective chances?

Ans18. Probability of A winning =  1/4
Probability of B winning = Probability of A losing * Probability of B winning = ¾ X 4/4 = 3/16
Probability of C winning = Probability of B losing * Probability of C winning = 13/16 X 1/4 = 13/64

Q19.   A standard deck of 52 cards is shuffled and the cards are dealt face up one at a time until an ace appears. Show that the probability of getting the first ace on or before the ninth card is greater than 50%.

Ans19. The probability of getting no aces in the first nine cards is 3/8…… = 40/44 = 43/92

Now 43/92 is less than 1/2, therefore getting first ace will be 1 – no ace, so greater than 50%

Q20.   To the nearest percent, the probability that any one person selected at random was born on a Monday is 14 percent.  What is the probability, to the nearest percent, that of any seven persons chosen at random, exactly one was born on a Monday?

Ans20. The probability of getting a Monday is 1/7 and the probability of getting 6 “non-Mondays” is (6/7)⁶.  The number of ways this can happen is a combination of “7 picking 1,” i.e. 7C1 = 7.

Therefore,

P(exactly 1 Monday out of 7 people) = 7C1 * (1/7)1 *(6/7)6 = .3966, so 40 %

Q21.   Mr. Ryan kept two matchboxes, one in each pocket. Each box contained exactly n matches. Whenever he wanted a match he reached at random into one of his pockets. When he found that the box he picked was empty, what is the probability that the other one has exactly k matches (k = < n)?

Ans21. If k matches remain in the other box, then n–k matches have been selected from that box. Suppose Mr. Ryan attempts to select the (n + 1)st match from the box in his left pocket. Then a total of 2n – k + 1 selections have been made; thus we have 22n – k+1 ways in which the matches can be selected. Of these, 2n – kCn (where mCr = m!/[r! (m – r)!] is the number of ways of choosing r outcomes out of m possibilities, ignoring order) combinations are such that the (n + 1)st selection is from his left pocket.

Therefore the probability the professor will open an empty box from his left pocket is 2n – kCn/22n–k+1. Of course, there is an equal probability that he will open an empty box from his right pocket.

Therefore the probability that the other box currently contains k matches is 2n – kCn/22n-k.

Q22.   Which is more likely, to get at least one double six in 24 throws of a pair of dice or to get at least one six in 4 throws of a die?

Ans22. P (no double sixes in 24 throws) = (35/36)24 =  0.509

P (at least one double six) = 1 – 0.509 = 0.491

P (No sixes in four throws) = (5/6)4 = 0.482

            P (at least one six in four throws) = 1 – 0.482 = 0.518, this is more likely.