Question

The probability of getting two tails when two coins are tossed is -

Correct
option
is $\dfrac14$

The sample space when two coins are tossed = (H, H), (H, T), (T, H), (T, T) So, n(S) = 4 The event "E" of getting two tails (T, T) = 1 So, n(E) = 1 So, the probability of getting two tails, P (E) = $\dfrac{n(E)}{n(S)}=\dfrac14$

The sample space when two coins are tossed = (H, H), (H, T), (T, H), (T, T) So, n(S) = 4 The event "E" of getting two tails (T, T) = 1 So, n(E) = 1 So, the probability of getting two tails, P (E) = $\dfrac{n(E)}{n(S)}=\dfrac14$

Question

What is the probability of getting an even number when a die is thrown?

Correct
option
is $\dfrac12$

The sample space when a dice is rolled, S = (1, 2, 3, 4, 5, and 6) So, n(S) = 6 E is the event of getting an even number. So, n (E) = 3 Probability of getting an even number P (E) = Total number of favorable outcomes/Total number of outcomes $\dfrac{n(E)}{n(S) }$$=\dfrac36=\dfrac12$

The sample space when a dice is rolled, S = (1, 2, 3, 4, 5, and 6) So, n(S) = 6 E is the event of getting an even number. So, n (E) = 3 Probability of getting an even number P (E) = Total number of favorable outcomes/Total number of outcomes $\dfrac{n(E)}{n(S) }$$=\dfrac36=\dfrac12$

Question

What is the probability of getting atleast one head if three unbiased coins are tossed?

Correct
option
is $\dfrac78$

The sample space is = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let E is the event of getting atleast one head Then E = {TTH, THT, HTT, THH, HTH, HHT, HHH} P(E)$ = \dfrac{n(E)} {n(S)} = \dfrac78$

The sample space is = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let E is the event of getting atleast one head Then E = {TTH, THT, HTT, THH, HTH, HHT, HHH} P(E)$ = \dfrac{n(E)} {n(S)} = \dfrac78$

Question

What is the probability of getting the sum as a prime number if two dice are thrown?

Correct
option
is $\dfrac5{12}$

As per the question: $n (S) = 6\times6 = 36$ And, the event that the sum is a prime number: $E = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}$ So, $n (E) = 15$ $\dfrac{n(E)}{n(S)} = 15/36 = 5/12$

As per the question: $n (S) = 6\times6 = 36$ And, the event that the sum is a prime number: $E = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}$ So, $n (E) = 15$ $\dfrac{n(E)}{n(S)} = 15/36 = 5/12$

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