The probability of an event of a trial
Answer: B
Every probability satisfies
\[ 0 \leq P(E) \leq 1 \]
with 0 for an impossible event and 1 for a certain one. Both bounds are attainable, so the interval is inclusive, and no probability can exceed 1.
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What is the probability of getting all heads or all tails, when three coins are tossed simultaneously ?
Answer: C
When three coins are tossed simultaneously, there are 8 possible outcomes, which are (HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT), where H is represents the head and T represents the tail.
Favorable outcomes of getting all heads or all tails are HHH and TTT.
We know that, Probability of an event E, P (E) =Number of favorable outcomes /Total number of outcomes
Hence, the required probability = 2/8 = 1/4
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There are 5 green, 6 black and 7 white balls in a bag. A ball is drawn at random from the bag. Find the probability that it is not white.
Answer: D
Given,
Number of green balls = 5
Number of black balls = 6
Number of white balls = 7
Total number of outcomes = 5 + 6 + 7 = 18
There are 18 balls out of which 11 are not white.
⇒ Number of favourable outcomes = 11
Probability of an event, P (E) = =Number of favourable outcomes / Total number of outcomes
⇒ P (ball drawn is not white) = 11/18
Therefore, probability that the ball drawn is not white is 11/18.
Alternate Method:
P (ball drawn is white) = 7/18
We know that,
P (ball drawn is white) + P (ball drawn is not white) = 1
Because the sum of the probability of an event and its complementary event is always 1.
⇒ P (ball drawn is not white) = 1 - P (ball drawn is white) = 1−7/18 = 11/18
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From a set of 17 cards, numbered 1, 2, …. 17, one card is drawn. What is the probability that the number is a multiple of 3 or 7?
Answer: B
The total number of possible outcomes is 17
The number of favourable outcomes is 3, 6, 7, 9, 12, 14 and 15
= 7
Thus, the required probability = number of favourable outcomes/ total number of outcomes
= 7/17.
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