Possibility : Test
17 min readJan 16, 2021
- If the probability of rain on any given day in City X is 50 percent, what is the probability that it rains on exactly 3 days in a 5-day period?
(A) 8/125
(B) 2/25
© 5/16
(D) 8/25
(E) 3/4 - If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?
A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32 - The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails?
(A) 1/8
(B) 1/2
© 3/4
(D) 7/8
(E) 15/16 - A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?
A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8 - A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?
A. 1/14
B. 1/7
C. 2/7
D. 3/7
E. 1/2 - An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ?
A. 1/9
B. 1/3
C. 1/2
D. 2/3
E. 5/6 - Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36 - A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses?
A. 2/16
B. 1/4
C. 7/24
D. 5/16
E. 15/32 - Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8 - Each of the marbles in a jar is either red or white or blue. If one marble is to be selected at random from the jar, what is the probability that the marble will be blue?
(1) There are a total of 24 marbles in the jar, 8 of which are red.
(2) The probability that the marble selected will be white is 1/2. - In a certain class, one student is to be selected at random to read. What is the probability that a boy will read?
(1) Two-thirds of the students in the class are boys.
(2) Ten of the students in the class are girls. - A jar contains 30 marbles, of which 20 are red and 10 are blue. If 9 of the marbles are removed, how many of the marbles left in the jar are red?
(1) Of the marbles removed, the ratio of the number of red ones to the number of blue ones is 2:1.
(2) Of the first 6 marbles removed, 4 are red. - What is the probability that a student randomly selected from a class of 60 students will be a male who has brown hair?
(1) One-half of the students have brown hair.
(2) One-third of the students are males. - A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15. - Terry holds 12 cards, each of which is red, white, green, or blue. If a person is to select a card randomly from the cards Terry is holding, is the probability less than 1/2 that the card selected will be either red or white?
(1) The probability that the person will select a blue card is 1/3
(2) The probability that the person will select a red card is 1/6 - Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?
(1) The probability that the ball will both be white and have an even number painted on it is 0
(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 - A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?
(1) y ≤ 8
(2) y ≥ 4 - A drawer contains 8 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?
(1) The probability is less than 0.2 that the first sock is black.
(2) The probability is more than 0.8 that the first sock is white. - There is a 50% chance Jen will visit Chile this year, while there is a 25% chance that she will visit Madagascar this year. What is the probability that Jen will visit either Chile or Madagascar this year, but NOT both?
A. 25%
B. 50%
C. 62.5%
D. 63.5%
E. 75% - First 15 positive integers are written on a board. If two numbers are selected one by one from the board at random (the numbers are not necessarily different), what is the probability that the sum of these numbers is odd?
- There is a group of people consisting of 10 men and 6 women. Among these 16 people, there are 4 married couples (man-woman couples). If one man and one woman are selected at random, what is the probability that a married couple gets selected?
- Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b?
(A) 5
(B) 6
© 11
(D) 16
(E) 19 - Alex tosses a coin four times. On two of the tosses (we don’t know which two), he gets ‘Heads’. What is the probability that he gets ‘Tails’ on other two tosses?
- Alex has five children. He has at least two girls (you do not know which two of her five children are girls). What is the probability that he has at least two boys too? (The probability of having a boy is 0.4 while the probability of having a girl is 0.6)
- On three consecutive flips of a coin, what is the probability that all three produce the same result?
(A) 1/16
(B) 1/8
© 1/4
(D) 3/8
(E) 1/2 - Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A) 66
B) 67
C) 68
D) 69
E) 70 - Two couples and one single person occupy a row of five chairs at random. What is the probability that neither couple sits together (the husband and the wife should not occupy adjacent seats)?
(A) 1/5
(B) 1/3
© 3/8
(D) 2/5
(E) 1/2 - Martin and Joey are playing a coin game in which each player tosses a fair coin alternately. The player who gets a ‘Heads’ first wins. The maximum number of tosses allowed in a single game for any player is 6. What is the probability that the person who tosses first will win the game?
- For one toss of a certain coin, the probability that the outcome is heads is 0.7. If the coin is tossed 6 times, what is the probability that the outcome will be tails at least 5 times?
- Six friends live in the city of Monrovia. There are four natural attractions around Monrovia — a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each one of them votes for the safari?
- Six friends live in the city of Monrovia. There are four natural attractions around Monrovia — a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each one of them votes for the same attraction?
- Six friends live in the city of Monrovia. There are four natural attractions around Monrovia — a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient
© Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
(D) EACH statement ALONE is sufficient
(E) Statements (1) and (2) TOGETHER are NOT sufficient
___________________________________________________________________
- C. the case of RRRNN can occur in many ways: RRRNN, RRNRRN, RNRRN, NRRRN, … basically it will be equal to # of arrangements (permutations) of 5 letters RRRNN out of which there are 3 identical R’s and 2 identical N’s. That # of arrangements is 5!/3!2!, (notice that it’s the same as 3C5). So, finally P=5!/3!2!∗(1/2)^5= 5/16
- E. We want the probability that coin will land heads up on the first 3 flips and not on the last 2 flips. So there is ONLY one favorable outcome, namely heads up on the first 3 flips and tails up on the last 2 flips: HHHTT.
# of total out comes is 2⁵=32.
P=favorable/total=1/32. - D.P(at last 1 tails) = 1 — P(all heads) = 1 — (1/2)³ = 7/8.
- D. Let’s find the probability of the opposite event and subtract this value from 1. The opposite event would be getting zero tails (so all heads) or 1 tail.
P(HHHH)=(12)4=1/16
P(THHH)=4!3!∗(12)4=4/16
we are multiplying by 4!/3! since THHH scenario can occur in number of ways: THHH, HTHH , HHTH, or HHHT (notice that 4!/3! basically gives number of arrangements of 4 letters THHH out of which 3 H’s are identcal).P(T≥2)=1−(116+416)=11/16 - D.(5C2 X 3C2) / 8C4
- D. n(n+1) to be divisible by 3 either n or n+1 must be a multiples of 3.
In each following group of numbers: {1, 2, 3}, {4, 5, 6}, {7, 8, 9}, …, {97, 98, 99} there are EXACTLY 2 numbers out of 3 satisfying the above condition. For example in {1, 2, 3} n can be: 2, or 3. Thus, the overall probability is 2/3. - D. half of the time the first dice will have even no. so prob will be greater than 1/2 as some cases extra when sum is 8.
so A,B,C ruled out
E gives 5 cases for sum 8.
but we know that sum cases will be common with “even condition” so D - At least 3 heads means 3, 4, or 5 heads.
3 consecutive heads
5 cases:HHHTT,THHHT,TTHHH,HTHHH,HHHTH
P=5∗(12)5=5/32P=5∗(1/2)^5=5/32.
4 consecutive heads
2 cases:HHHHT,THHHH
P=2∗(12)5=2/32P=2∗(1/2)^5=232.
5 consecutive heads
1 case:HHHHH
P=(12)5=1/32P=(1/2)^5=1/32.
P=5/32+2/32+1/32=8/32=1/4 - D.Total # of ways of assigning 4 letters to 4 envelopes is 4!=244!=24.
Only one letter in the right envelope: 4(# of envelopes)*2(# of ways possible to arrange 3 letters incorrectly in the envelopes, when one is correct).
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
As we have 4 letters, total # of ways 4*2=8
= 8/24 = 1/3 - C.
- A.
- A.
- E.
- D.(1) The probability that the two bulbs to be drawn will be defective is 1/15 → clearly sufficient, as probability, pp, of drawing 2 defective bulbs out of total 10 bulbs, obviously depends on # of defective bulbs, nn, so we can calculate uniques value of nn if we are given pp.
To show how it can be done: n10∗n−19=115n10∗n−19=115 → n(n−1)=6n(n−1)=6 → n=3n=3 or n=−2n=−2 (not a valid solution as nn represents # of defective bulbs and can not be negative). Sufficient.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15 → also sufficient, but a little bit trickier: if it were 3 defective and 7 good bulbs OR 7 defective and 3 good bulbs, then the probability of drawing one defective and one good bulb would be the same for both cases (symmetric distribution), so info about the probability, 7/15, of drawing one defective and one good bulb would give us 2 values of nn one less than 5 and another more than 5 (their sum would be 10), but as we are given that n<5n<5, we can stiil get unique value of nn which is less than 5.
To show how it can be done: 2∗n10∗10−n9=7152∗n10∗10−n9=715 → n(10−n)=21n(10−n)=21 → n=3n=3 or n=7n=7 (not a valid solution as n<5n<5). Sufficient. - E.
- E. (1) The probability that the ball will both be white and have an even number painted on it is 0 →P(W&E)=0 (no white ball with even number) → P(WorE)=P(W)+P(E)−0P(WorE)=P(W)+P(E)−0. Not sufficient
(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 → P(W)−P(E)=0.2P(W)−P(E)=0.2, multiple values are possible for P(W)P(W) and P(E)P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE)P(WorE)
(1)+(2)
P(W&E)=0 and P(W)−P(E)=0.2P(W)−P(E)=0.2 → P(WorE)=2P(E)+0.2P(WorE)=2P(E)+0.2 → multiple answers are possible, for instance: if P(E)=0.4P(E)=0.4 (10 even balls) then P(WorE)=1P(WorE)=1 BUT if P(E)=0.2P(E)=0.2 (5 even balls) then P(WorE)=0.6P(WorE)=0.6. Not sufficient. - B. # of total marbles in the jar equals to 8+y8+y, out of which R=8R=8 and W=yW=y
P(RR)=88+y∗8−18+y−1=88+y∗77+yP(RR)=88+y∗8−18+y−1=88+y∗77+y;
P(RW)=2∗88+y∗y8+y−1=2∗88+y∗y7+yP(RW)=2∗88+y∗y8+y−1=2∗88+y∗y7+y, multiplying by 2 as RW can occur in two ways RW or WR;
Question: is 88+y∗77+y>288+y∗y7+y88+y∗77+y>288+y∗y7+y? → is 72>y72>y? → is y❤.5y❤.5? eg 0, 1, 2, 3.
(1) y ≤ 8, not sufficient.
(2) y ≥ 4, sufficient, (yy is not less than 3.5). - D.
(1) B8<0.2B8<0.2 → B<1.6B<1.6, so there can be 1 or 0 black socks in the drawer. In any case as the # is less then 2 the probability of picking 2 black socks is 0. Sufficient.
(2) W8>0.8W8>0.8 → W>6.4W>6.4, so there are 7 or 8 white socks in the drawer. As the maximum possible # of black socks is 1, thus the probability of 2 blacks is 0. Sufficient. - B. In practice, this question could then be solved using the default formula of P(C) + P(M) — P(C&M) and then subtracting P(C&M) again… or simply P(C) + P(M) — 2*P(C&M). Simply put, the first two arguments in the formula account for the “both” possibility twice, so we must remove it twice to answer the question at hand.
A somewhat similar yet more straight forward alternative is to go bottom up instead of top down. The probability of going only to Chile is P(C) * P(¬M) = 0.5 * 0.75 = 0.375. The probability of going only to Madagascar is P(M) * P(¬C) = 0.25 * 0.5 = 0.125. Adding those two probabilities together yields the correct answer of 0.5 or 50%. The revised formula also yields this result (0.5 + 0.25–2 (0.125) = 0.5), so we can feel confident in our answer choice and not any of the other tempting answer choices. - Numbers: 1, 2, 3, 4, …, 13, 14, 15
When will the sum of two of these numbers be odd? When one number is odd and the other is even.
P(Sum is Odd) = P(First number is Odd)*P(Second number is even) + P(First number is Even)*P(Second number is Odd)
P(Sum is Odd) = (8/15)*(7/15) + (7/15)*(8/15) = 112/225 - P(Selecting a Married Man) = 4/10
P(Selecting the Wife of that Man) = 1/6
P(Married Couple is Selected) = (4/10)*(1/6) = 4/60 - E.The first element can be chosen in 4 ways — one of 3, 4, 5 and 6. (You are given that 2 cannot be the first element).The second element can be chosen in 4 ways (2 and the leftover 3 numbers).The third element can be chosen in 3 ways.The fourth element can be chosen in 2 ways.And finally there will be only 1 element left for the last spot.
Number of ways of making set S = 4*4*3*2*1 = 96
In how many of these sets will 5 be in the second spot?
If 5 is reserved for the second spot, there are only 3 ways of filling the first spot (3 or 4 or 6).The second spot has to be taken by 5.The third element will be chosen in 3 ways (ignoring 5 and the first spot)The fourth element can be chosen in 2 ways.And finally there will be only 1 element left for the last spot.
Number of favorable cases = 3*1*3*2*1 = 18
Required Probability = Favorable Cases/Total Cases = 18/96 = 3/16 = a/b
a+b = 3 + 16 = 19 - We can easily find P(2 Heads and 2 Tails) and P(At least 2 Heads) since we are comfortable with the concepts of binomial probability! (right?)
P(2 Heads and 2 Tails) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/(2!*2!) = 3/8
You multiply by 4!/(2!*2!) because out of the four tosses, any 2 could be heads and the other two would be tails. So you have to account for all arrangements: HHTT, HTHT, TTHH etc
P(Atleast 2 Heads) = P(2 Heads and 2 Tails) + P(3 Heads, 1 Tails) + P(4 Heads)
Let me remind you here that we can also find P(Atleast 2 Heads) in the reverse way like this:
P(Atleast 2 Heads) = 1 — [P(4 Tails) + P(3 Tails, 1 Heads)]
Let me show you the calculations involved in both the methods.
P(2 Heads and 2 Tails) = 3/8 (calculated above)
P(3 Heads, 1 Tails) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/3! = 1/4
We multiply by 4!/3! to account for all arrangements e.g. HHHT, HHTH etc
P(4 Heads) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16
P(Atleast 2 Heads) = 3/8 + 1/4 + 1/16 = 11/16 - Method 1:
P(At least 2 Girls) = P(2 Girls and 3 Boys) + P(3 Girls and 2 Boys) + P(4 Girls + 1 Boy) + P(5 Girls)
P(2 Girls and 3 Boys)=(0.4)3∗(0.6)2∗10P(2 Girls and 3 Boys)=(0.4)3∗(0.6)2∗10 (from above)
P(3 Girls and 2 Boys)=(0.4)2∗(0.6)3∗10P(3 Girls and 2 Boys)=(0.4)2∗(0.6)3∗10 (from above)
P(4 Girls + 1 Boy)=(0.4)∗(0.6)∗(0.6)∗(0.6)∗(0.6)∗5!/4!=(0.4)∗(0.6)4∗5P(4 Girls + 1 Boy)=(0.4)∗(0.6)∗(0.6)∗(0.6)∗(0.6)∗5!/4!=(0.4)∗(0.6)4∗5
P(5 Girls)=(0.6)∗(0.6)∗(0.6)∗(0.6)∗(0.6)=(0.6)5P(5 Girls)=(0.6)∗(0.6)∗(0.6)∗(0.6)∗(0.6)=(0.6)5
P(At least 2 Girls)=[(0.4)3∗(0.6)2∗10]+[(0.4)2∗(0.6)3∗10]+[(0.4)∗(0.6)4∗5]+[(0.6)5]P(At least 2 Girls)=[(0.4)3∗(0.6)2∗10]+[(0.4)2∗(0.6)3∗10]+[(0.4)∗(0.6)4∗5]+[(0.6)5]
Method 2:
P(At least 2 Girls)=1–P(5 Boys)–P(1 Girl and 4 Boys)P(At least 2 Girls)=1–P(5 Boys)–P(1 Girl and 4 Boys)
P(5 Boys)=(0.4)∗(0.4)∗(0.4)∗(0.4)∗(0.4)=(0.4)5P(5 Boys)=(0.4)∗(0.4)∗(0.4)∗(0.4)∗(0.4)=(0.4)5
P(1 Girl and 4 Boys)=(0.6)∗(0.4)∗(0.4)∗(0.4)∗(0.4)∗5!4!=(0.6)∗(0.4)4∗5P(1 Girl and 4 Boys)=(0.6)∗(0.4)∗(0.4)∗(0.4)∗(0.4)∗5!4!=(0.6)∗(0.4)4∗5
P(At least 2 Girls)=1–[(0.4)5]–[(0.6)∗(0.4)4∗5] - C. this case the result of the first flip doesn’t have to be one or the other. Your job is just to match whatever the first result was on the next two. If the first was heads, then you need heads next (a 1/2 chance) and heads again (a 1/2 chance). And if it were tails, then you need tails (1/2) then tails (1/2). But because “any match will do” and you don’t care that it’s a specific match — the question doesn’t specify all heads or all tails, just “all of one of them” — your probability doubles because you’re not concerned about the result of the first event, you’re only concerned about matching whatever that result was.
- A. With 2 and 5 cent coins as your options, you can’t get to 1 and you can’t get to 3, so those are two “ain’t one” possibilities. And then “100% minus… comes back into play” — Notice too that 70¢ is the maximum possible sum (that would use all the coins), so 70¢ — 1¢, or 69¢, and 70¢ — 3¢, or 67¢ are impossible too. So the answer is 66, but the takeaway is bigger: when calculating all the possibilities looks to be far too time-consuming, you often have the opportunity to calculate the possibilities that “ain’t one.” You’ve got a lot of problems to tackle on test day; hopefully this strategy allows you to make one question much less of one.
- D.Number of arrangements in which both couples sit together: Let’s say the two couples are {C1h, C1w} and {C2h, C2w} and the single person is S. There are three groups/individuals. They can be arranged in 3! ways. But in each couple, husband and wife can be arranged in 2 ways (husband and wife can switch places)
Hence, number of arrangements such that both couples are together = 3!*2*2 = 24
Number of arrangements such that C1h and C1w are together: C1 acts as one group. We can arrange 4 people/groups in 4! ways. C1h and C1w can be arranged in 2 ways (husband and wife can switch places).
Number of arrangements in which C1h and C1w are together = 4! * 2 = 48
But this 48 includes the number of arrangements in which C2h and C2w are also sitting together.
Therefore, number of arrangements such that ONLY C1h and C1w sit together = 48–24 = 24
Similarly, number of arrangements such that ONLY C2h and C2w sit together = 24
Number of arrangements in which at least one couple sits together = 24 + 24 + 24 = 72
Number of arrangements in which neither couple sits together = 120–72 = 48
Probability that neither couple sits together = 48/120 = 2/5 - In the worst case, the first person will have to toss 6 times to get a ‘Heads’. He and the second person would end up getting ‘Tails’ on five previous tosses.
Probability that the first person tosses the coin and gets a ‘Heads’ right away = 1/2
Probability that the first person tosses the coin and gets a ‘Tails’ (1/2), the second person gets ‘Tails’ (1/2) and then the first person gets a ‘Heads’(1/2) = (1/2)*(1/2)*(1/2) = (1/2)³
Probability that the first person tosses the coin and gets a ‘Tails’ (1/2), the second person gets ‘Tails’ (1/2) , the first person tosses again and gets a ‘Tails’ again (1/2), the second person gets ‘Tails’ again (1/2) and finally, the first person tosses and this time, gets a ‘Heads’ (1/2) = (1/2)⁵
and so on…
Probability that the first person will have to toss 6 times to get a ‘Heads’ = (1/2)¹¹
Probability that the first person will win = (1/2) + (1/2)³ + (1/2)⁵ + (1/2)⁷ + (1/2)⁹ + (1/2)¹¹ - The only acceptable cases are those in which we get ‘tails’ on all 6 tosses or we get tails on exactly 5 of the 6 tosses.
P(Tails on all 6 tosses) = (0.3)∗(0.3)∗(0.3)∗(0.3)∗(0.3)∗(0.3)=(0.3)^6
P(Tails on exactly 5 tosses and Heads on one toss) = (0.3)5∗(0.7)∗6
We multiply by 6 because 5 tails and 1 heads can be obtained in 6 different ways: HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, TTTTTH
Probability that the outcome will be tails at least 5 times = (0.3)^6+(0.3)^5∗(0.7)∗6 - What is the no. of ways in which the friends can vote? Say, the friends are A, B, C, D, E and F. A can vote in 4 ways. B can vote in 4 ways. C can vote in 4 ways and so on… Total no of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4⁶ (Using our old friend, the basic counting principle). We discussed this concept in our post on Unfair Distributions.
Therefore, P(A) = 1(4^6) - P(A) = 4/(4^6)=1/(4^5)
Case 1: (1, 1, 1, 3)
First, we select the attraction that will get 3 votes in 4 ways (= 4C1)
Now, we can select the 3 people who will vote for this attraction in 6*5*4/3! = 20 ways (= 6C3 )
The other 3 votes will be distributed among the other 3 attractions in 3! = 6 ways
The 6 people could vote for the 4 attractions in this case in 4*20*6 = 480 ways
Case 2: (1, 1, 2, 2)
Let’s select the two attractions that will get 2 votes each in 4*3/2! = 6 ways (= 4C2). Say we select caves and waterfall.
Now, we can select the 2 people who will vote for one of the selected attractions in 6*5/2! = 15 ways (= 6C2)
We can select the other 2 people who will vote for the other selected attraction in 4*3/2! = 6 ways (= 4C2)
The other 2 votes will be distributed among the other 2 attractions in 2! = 2 ways
The 6 people could vote for the 4 attractions in this case in 6*15*6*2 = 1080 ways
Total number of ways in which 6 votes can be distributed among 4 attractions such that each attraction gets at least one vote = 480 + 1080 = 1560 ways
As we saw in the questions above, the total no. of ways in which the friends can vote = 4⁶
Therefore, P(A) = 1560(46)
