Probability - Study Mode
[#91] Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
Correct Answer
(B) $$frac{{3}}{{4}}$$
Explanation
Solution: In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36 Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} $$eqalign{
& herefore nleft( E
ight) = 27 cr
& herefore Pleft( E
ight) = frac{{nleft( E
ight)}}{{nleft( S
ight)}} = frac{{27}}{{36}} = frac{3}{4} cr} $$
[#92] In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
Correct Answer
(A) $$frac{{21}}{{46}}$$
Explanation
Solution: Let S be the sample space and E be the event of selecting 1 girl and 2 boys Then, n(S) = Number ways of selecting 3 students out of 25 $$eqalign{
& = {}^{25}{C_3} cr
& = frac{{ {25 imes 24 imes 23} }}{{ {3 imes 2 imes 1} }} cr
& = 2300 cr
& nleft( E
ight) = {^{10}{C_1}{ imes ^{15}}{C_2}} cr
& = {10 imes frac{{ {15 imes 14} }}{{ {2 imes 1} }}} cr
& = 1050 cr
& herefore Pleft( E
ight) = frac{{nleft( E
ight)}}{{nleft( S
ight)}} = frac{{1050}}{{2300}} = frac{{21}}{{46}} cr} $$
[#93] In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
Correct Answer
(C) $$frac{{2}}{{7}}$$
Explanation
Solution: $$eqalign{
& Pleft( {{ ext{getting}},{ ext{a}},{ ext{prize}}}
ight) cr
& = frac{{10}}{{10 + 25}} cr
& = frac{{10}}{{35}} cr
& = frac{2}{7} cr} $$
[#94] Two dice are tossed. The probability that the total score is a prime number is:
Correct Answer
(B) $$frac{{5}}{{12}}$$
Explanation
Solution: Clearly, n(S) = (6 x 6) = 36 Let E = Event that the sum is a prime number.Then E = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)} $$eqalign{
& herefore nleft( E
ight) = 15 cr
& herefore Pleft( E
ight) = frac{{nleft( E
ight)}}{{nleft( S
ight)}} cr
& = frac{{15}}{{36}} cr
& = frac{5}{{12}} cr} $$ Alternate solution Understanding the Question: We're tossing two dice. Each die has six sides (1, 2, 3, 4, 5, 6). We want to find the probability that the sum of the numbers shown on both dice is a prime number. A prime number is a number greater than 1 that is only divisible by 1 and itself (e.g., 2, 3, 5, 7, 11...). Finding the Possible Outcomes: First, let's figure out all the possible sums we can get when we add the numbers on two dice. The minimum sum is 2 (1+1) and the maximum is 12 (6+6). Identifying Prime Number Sums: Now, let's list the sums that are prime numbers: 2, 3, 5, 7, 11. Counting Favorable Outcomes: Let's count how many ways we can get each of these prime sums: * 2: Only one way (1+1) * 3: Two ways (1+2, 2+1) * 5: Four ways (1+4, 2+3, 3+2, 4+1) * 7: Six ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) * 11: Two ways (5+6, 6+5) Total favorable outcomes (prime number sums): 1 + 2 + 4 + 6 + 2 = 15 Calculating Total Possible Outcomes: The total number of possible outcomes when tossing two dice is 6 (outcomes for the first die) * 6 (outcomes for the second die) = 36 Calculating Probability: Probability is calculated as (Favorable Outcomes) / (Total Possible Outcomes). So, the probability of getting a prime number sum is 15/36. This simplifies to 5/12. Therefore, the correct option is B: 5/12
[#95] Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is at least one graduate among them?
Correct Answer
(D) $$frac{{5}}{{6}}$$
Explanation
Solution: P(at least one graduate) = 1 - P(no graduates) $$eqalign{
& = 1 - frac{{{}^6{C_3}}}{{{}^{10}{C_3}}} cr
& = 1 - frac{{6 imes 5 imes 4}}{{10 imes 9 imes 8}} cr
& = frac{5}{6} cr} $$