- A dice is rolled three times and sum of three numbers appearing
- A die is rolled three times. The probability of getting a larger number than the previous number each time is
- Probability Distributions for Discrete Random Variables
- A die is rolled three times. The probability of getting a larger number than the previous number each time is
- Probability Distributions for Discrete Random Variables
- A dice is rolled three times and sum of three numbers appearing
- Probability Distributions for Discrete Random Variables
- A dice is rolled three times and sum of three numbers appearing
- A die is rolled three times. The probability of getting a larger number than the previous number each time is
- A dice is rolled three times and sum of three numbers appearing
A dice is rolled three times and sum of three numbers appearing
Correct Option: D Total number of favourable outcomes n(S) = 6 3 = 216 Combinations of outcomes for getting sum of 15 on uppermost face = (4, 5, 6), (5, 4, 6), ( 6, 5, 4), (5, 6, 4), (4, 6, 5), (6, 4, 5), (5, 5, 5),(6, 6, 3), (6, 3, 6), (3, 6, 6) Now, outcomes on which first roll was a four, n(E) = (4, 5, 6),(4, 6, 5) ∴ P(E) = n(E)/n(S) = 2/216 =1/108
A die is rolled three times. The probability of getting a larger number than the previous number each time is
The correct option is B 5 54 Find the probability of getting a larger number than the previous number each time. According to the question, we have n = 6 , r = 3 Total number of sample space when die is rolled 3 times = 6 3 [ ∵ Total number of sample space when die is rolled r times = n r ] Number of favourable outcomes = C 3 6 Therefore, the required probability = C 3 6 6 3 [ ∵ P r o b a b i l i t y = N u m b e r o f f a v o u r a b l e o u t c o m e s t o t a l n u m b e r o f o u t c o m e s ] = 6 ! 6 - 3 ! 3 ! 216 = 5 54 Hence, Option B is the correct answer.
Probability Distributions for Discrete Random Variables
Definition The A list of each possible value and its probability. of a discrete random variable X is a list of each possible value of X together with the probability that X takes that value in one trial of the experiment. The probabilities in the probability distribution of a random variable X must satisfy the following two conditions: Example 1 A fair coin is tossed twice. Let X be the number of heads that are observed. • Construct the probability distribution of X. • Find the probability that at least one head is observed. Solution: • The possible values that X can take are 0, 1, and 2. Each of these numbers corresponds to an event in the sample space S = . The probability of each of these events, hence of the corresponding value of X, can be found simply by counting, to give x 0 1 2 P ( x ) 0.25 0.50 0.25 This table is the probability distribution of X. • “At least one head” is the event X ≥ 1, which is the union of the mutually exclusive events X = 1 and X = 2. Thus P ( X ≥ 1 ) = P ( 1 ) + P ( 2 ) = 0.50 + 0.25 = 0.75 A histogram that graphically illustrates the probability distribution is given in Example 2 A pair of fair dice is rolled. Let X denote the sum of the number of dots on the top faces. • Construct the probability distribution of X. • Find P( X ≥ 9). • Find the probability that X takes an even value. Solution: The sample space of equally likely outcomes is 11 12 13 14 15 16 21 22 23 24 25 26 31 32 33 34 35 36 41 42 43 44 45 46 51 52 53 54 55 56 61 62 63 64...
A die is rolled three times. The probability of getting a larger number than the previous number each time is
The correct option is B 5 54 Find the probability of getting a larger number than the previous number each time. According to the question, we have n = 6 , r = 3 Total number of sample space when die is rolled 3 times = 6 3 [ ∵ Total number of sample space when die is rolled r times = n r ] Number of favourable outcomes = C 3 6 Therefore, the required probability = C 3 6 6 3 [ ∵ P r o b a b i l i t y = N u m b e r o f f a v o u r a b l e o u t c o m e s t o t a l n u m b e r o f o u t c o m e s ] = 6 ! 6 - 3 ! 3 ! 216 = 5 54 Hence, Option B is the correct answer.
Probability Distributions for Discrete Random Variables
Definition The A list of each possible value and its probability. of a discrete random variable X is a list of each possible value of X together with the probability that X takes that value in one trial of the experiment. The probabilities in the probability distribution of a random variable X must satisfy the following two conditions: Example 1 A fair coin is tossed twice. Let X be the number of heads that are observed. • Construct the probability distribution of X. • Find the probability that at least one head is observed. Solution: • The possible values that X can take are 0, 1, and 2. Each of these numbers corresponds to an event in the sample space S = . The probability of each of these events, hence of the corresponding value of X, can be found simply by counting, to give x 0 1 2 P ( x ) 0.25 0.50 0.25 This table is the probability distribution of X. • “At least one head” is the event X ≥ 1, which is the union of the mutually exclusive events X = 1 and X = 2. Thus P ( X ≥ 1 ) = P ( 1 ) + P ( 2 ) = 0.50 + 0.25 = 0.75 A histogram that graphically illustrates the probability distribution is given in Example 2 A pair of fair dice is rolled. Let X denote the sum of the number of dots on the top faces. • Construct the probability distribution of X. • Find P( X ≥ 9). • Find the probability that X takes an even value. Solution: The sample space of equally likely outcomes is 11 12 13 14 15 16 21 22 23 24 25 26 31 32 33 34 35 36 41 42 43 44 45 46 51 52 53 54 55 56 61 62 63 64...
A dice is rolled three times and sum of three numbers appearing
Correct Option: D Total number of favourable outcomes n(S) = 6 3 = 216 Combinations of outcomes for getting sum of 15 on uppermost face = (4, 5, 6), (5, 4, 6), ( 6, 5, 4), (5, 6, 4), (4, 6, 5), (6, 4, 5), (5, 5, 5),(6, 6, 3), (6, 3, 6), (3, 6, 6) Now, outcomes on which first roll was a four, n(E) = (4, 5, 6),(4, 6, 5) ∴ P(E) = n(E)/n(S) = 2/216 =1/108
Probability Distributions for Discrete Random Variables
Definition The A list of each possible value and its probability. of a discrete random variable X is a list of each possible value of X together with the probability that X takes that value in one trial of the experiment. The probabilities in the probability distribution of a random variable X must satisfy the following two conditions: Example 1 A fair coin is tossed twice. Let X be the number of heads that are observed. • Construct the probability distribution of X. • Find the probability that at least one head is observed. Solution: • The possible values that X can take are 0, 1, and 2. Each of these numbers corresponds to an event in the sample space S = . The probability of each of these events, hence of the corresponding value of X, can be found simply by counting, to give x 0 1 2 P ( x ) 0.25 0.50 0.25 This table is the probability distribution of X. • “At least one head” is the event X ≥ 1, which is the union of the mutually exclusive events X = 1 and X = 2. Thus P ( X ≥ 1 ) = P ( 1 ) + P ( 2 ) = 0.50 + 0.25 = 0.75 A histogram that graphically illustrates the probability distribution is given in Example 2 A pair of fair dice is rolled. Let X denote the sum of the number of dots on the top faces. • Construct the probability distribution of X. • Find P( X ≥ 9). • Find the probability that X takes an even value. Solution: The sample space of equally likely outcomes is 11 12 13 14 15 16 21 22 23 24 25 26 31 32 33 34 35 36 41 42 43 44 45 46 51 52 53 54 55 56 61 62 63 64...
A dice is rolled three times and sum of three numbers appearing
Correct Option: D Total number of favourable outcomes n(S) = 6 3 = 216 Combinations of outcomes for getting sum of 15 on uppermost face = (4, 5, 6), (5, 4, 6), ( 6, 5, 4), (5, 6, 4), (4, 6, 5), (6, 4, 5), (5, 5, 5),(6, 6, 3), (6, 3, 6), (3, 6, 6) Now, outcomes on which first roll was a four, n(E) = (4, 5, 6),(4, 6, 5) ∴ P(E) = n(E)/n(S) = 2/216 =1/108
A die is rolled three times. The probability of getting a larger number than the previous number each time is
The correct option is B 5 54 Find the probability of getting a larger number than the previous number each time. According to the question, we have n = 6 , r = 3 Total number of sample space when die is rolled 3 times = 6 3 [ ∵ Total number of sample space when die is rolled r times = n r ] Number of favourable outcomes = C 3 6 Therefore, the required probability = C 3 6 6 3 [ ∵ P r o b a b i l i t y = N u m b e r o f f a v o u r a b l e o u t c o m e s t o t a l n u m b e r o f o u t c o m e s ] = 6 ! 6 - 3 ! 3 ! 216 = 5 54 Hence, Option B is the correct answer.
A dice is rolled three times and sum of three numbers appearing
Correct Option: D Total number of favourable outcomes n(S) = 6 3 = 216 Combinations of outcomes for getting sum of 15 on uppermost face = (4, 5, 6), (5, 4, 6), ( 6, 5, 4), (5, 6, 4), (4, 6, 5), (6, 4, 5), (5, 5, 5),(6, 6, 3), (6, 3, 6), (3, 6, 6) Now, outcomes on which first roll was a four, n(E) = (4, 5, 6),(4, 6, 5) ∴ P(E) = n(E)/n(S) = 2/216 =1/108
