Class 10 Mathematics Chapter 14 Probability

Key Concept: Zero Probability

a) 0
[Solution Description]
A standard six-sided die only has numbers from 1 to 6. Thus, rolling a 7 is an impossible event.

Therefore, the probability of this event is 0.

The probability of an impossible event is always 0. For example, rolling a 7 on a standard six-sided die is impossible because the die only has numbers from 1 to 6. Therefore, the probability of this event is 0, illustrating the concept of an impossible event in probability.

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Key Concept: Conditional Probability

b) $\frac{1}{2}$
[Solution Description]
The total number of red cards is 26 (13 hearts and 13 diamonds). Since diamonds make up half of the red cards,
$
P(\text{Diamond | Red}) = \frac{\text{Number of Diamonds}}{\text{Total Number of Red Cards}} = \frac{13}{26} = \frac{1}{2}
$.

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Key Concept: Coin Toss

b) $\frac{1}{2}$
[Solution Description] When a fair coin is tossed, there are two possible outcomes: heads or tails. Therefore, the probability P(T) of getting tails is:
$
P(T) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{2}
$
So, the probability of getting tails is $\frac{1}{2}$.

When a fair coin is tossed, it is assumed to be unbiased, meaning it has no preference for landing on heads or tails. The possible outcomes are heads or tails, making them equally likely. The probability of getting tails is calculated by dividing the number of favorable outcomes (1 tail) by the total number of outcomes (2 possible outcomes).

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Key Concept: Evolution of Probability, Comparative Contributions

c) J.Cardan, James Bernoulli, A. de Moivre, Pierre Simon Laplace
[Solution Description] Chronologically, J.Cardan was first with his book on games of chance. Then came James Bernoulli, who made significant early advancements. Following him was A. de Moivre, notable for developing analytical methods. Lastly, Pierre Simon Laplace further refined probability theory and its application.

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Key Concept: Multiple Events

c) 0.105
[Solution Description]
The probability that event A does not occur is the complement of A, calculated as $P(\text{not A}) = 1 – P(A) = 1 – 0.65 = 0.35$. Since events A and B are unrelated, their combined probability when A does not occur is the product of their individual probabilities:

$
P(\text{not A and B}) = P(\text{not A}) \times P(B) = 0.35 \times 0.3 = 0.105
$

Thus, the probability that A does not occur and B occurs is 0.105.

When dealing with multiple events, the probability of one event not occurring while another unrelated event occurs can be calculated using the complementary probability formula. If events A and B are independent, the probability of event A not occurring is $ P(\text{not A}) = 1 – P(A) $. The combined probability of A not occurring and B occurring is the product of their individual probabilities.

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Key Concept: Probability of Combined Events

b) $\frac{4}{13}$
[Solution Description] The probability of drawing a king is:
$
P(\text{king}) = \frac{4}{52} = \frac{1}{13}.
$
The probability of drawing a diamond is:
$P(\text{diamond}) = \frac{13}{52} = \frac{1}{4}.$
There is an overlap because one king is also a diamond. Hence, using the inclusion-exclusion principle:
$
P(\text{king or diamond}) = P(\text{king}) + P(\text{diamond}) – P(\text{king and diamond}) = \frac{1}{13} + \frac{1}{4} – \frac{1}{52}.
$
Simplifying, we calculate:
$
P(\text{king or diamond}) = \frac{4}{52} + \frac{13}{52} – \frac{1}{52} = \frac{16}{52} =\ \frac{4}{13}.
$
Therefore, the probability is $\frac{4}{13}$.

The probability of drawing a card that is either a king or a diamond from a standard deck involves the concept of combined events and the inclusion-exclusion principle. The probability of drawing a king is $\frac{1}{13}$, and the probability of drawing a diamond is $\frac{1}{4}$. Since one card is both a king and a diamond, the probability of drawing either is calculated by adding the individual probabilities and subtracting the overlap.

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Key Concept: Complementary Events

c) $\frac{3}{5}$
[Solution Description]
The probability of the event not happening is given by the complementary probability formula:

$
P(\text{not raining}) = 1 – P(\text{raining}) = 1 – \frac{2}{5} = \frac{3}{5}
$

Therefore, the probability it will not rain is $\frac{3}{5}$.

The concept of complementary events in probability states that the probability of an event not occurring is equal to 1 minus the probability of the event occurring. If the probability of it raining tomorrow is $\frac{2}{5}$, then the probability of it not raining is calculated as $1 – \frac{2}{5}$.

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Key Concept: Two Dice Sum

b) $\frac{1}{9}$
[Solution Description]
To find this probability, we first determine all possible rolls that result in a sum of 9: $(3,6), (4,5), (5,4), (6,3)$. There are 4 such outcomes.

The total number of outcomes when two dice are rolled is $6 \times 6 = 36$.

Therefore, the probability is:

$
P(\text{sum is 9}) = \frac{4}{36} = \frac{1}{9}
$

When rolling two dice, the total number of possible outcomes is $6 \times 6 = 36$, as each die has 6 faces. To find the probability of a specific sum, such as 9, we identify all combinations of dice rolls that result in that sum. The probability is then calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

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Key Concept: Total Outcomes

c) 15
[Solution Description]
To find the total number of outcomes, sum up all the marbles:

Number of blue marbles = 4

Number of white marbles = 6

Number of black marbles = 5

Total outcomes = 4 (blue) + 6 (white) + 5 (black) = 15

There are 15 possible outcomes when drawing one marble.

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Key Concept: Probability of Specific Sum

b) $\frac{1}{6}$
[Solution Description] Possible combinations for a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 favorable outcomes. Total possible outcomes when rolling two dice is $6 \times 6 = 36$. Thus, $P(\text{sum } 7) = \frac{6}{36} = \frac{1}{6}$.

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Key Concept: Complementary Events

c) 0.7
[Solution Description] The probability of the complementary event is calculated as $P(\text{not } E) = 1 – P(E)$. Here, $P(\text{not } \text{raining}) = 1 – 0.3 = 0.7$.

Complementary events are pairs of events where the occurrence of one event means the other cannot occur. The probability of a complementary event is given by $P(\text{not } E) = 1 – P(E)$. For example, if the probability of raining is 0.3, the probability of it not raining is $1 – 0.3 = 0.7$.

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Key Concept: Coin Toss

b) $\frac{1}{2}$
[Solution Description] When two coins are tossed, possible outcomes are: HH, HT, TH, TT. Exactly one head occurs in HT or TH. Thus, there are 2 favorable outcomes.
$
P(\text{exactly one head}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{4} = \frac{1}{2}
$

When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), and (T, T). The probability of getting exactly one head involves the outcomes (H, T) and (T, H). The probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

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Key Concept: Real-World Application, Conditional Probability

c) 0.75
[Solution Description] Given that the first toss is a head, we need at least one more head in the remaining two tosses. Total possibilities for two tosses are $2^2 = 4$: HH, HT, TH, TT. To have at least two heads, valid sequences are: HH, HT, TH.

Count of sequences with at least one additional head (HH, HT, TH): 3

Thus, the probability becomes:

$
\text{Probability (at least two heads)} = \frac{3}{4} = 0.75
$

Therefore, the probability that the decision will proceed is 0.75.

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Key Concept: Complementary Events

c) 0.6
[Solution Description] The probability of complementary events satisfies $$P(A) + P(\text{not } A) = 1$$. Therefore, $$P(\text{not } A) = 1 – 0.4 = 0.6$$.

Complementary events are pairs of events where one event is the complement of the other. If the probability of event A is given as P(A), then the probability of the complement of A, denoted as $P(\text{not } A)$, is calculated using the formula $P(\text{not } A) = 1 – P(A)$. This relationship ensures that the sum of the probabilities of an event and its complement is always 1.

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Key Concept: Probability in Decision Making, Complementary in Risk Assessment

b) 0.30
[Solution Description] The probability that the product will not be successful can be found using complementary events: $P(\text{not successful}) = 1 – P(\text{successful})$. Substituting the given values:

$
P(\text{not successful}) = 1 – 0.7 = 0.3
$

The probability that the backup plan will be needed is therefore 0.3.

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Key Concept: Probability Calculation

b) $ \frac{1}{2} $
[Solution Description] There are 6 possible outcomes when a die is thrown: 1, 2, 3, 4, 5, and 6. The odd numbers among these are 1, 3, and 5. Therefore, there are 3 favorable outcomes for getting an odd number. According to the formula for theoretical probability:

$
P(\text{odd number}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{6} = \frac{1}{2}
$

Thus, the probability of getting an odd number is $\frac{1}{2}$.

The concept of equally likely outcomes is crucial in determining the probability of an event. When a six-sided die is thrown, each face (number) has an equal chance of landing face up. Therefore, the outcomes are equally likely. The theoretical probability of an event, such as getting an odd number, is calculated by dividing the number of favorable outcomes by the total number of possible outcomes, assuming all outcomes are equally likely.

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Key Concept: Basic Gender Probability

c) $ \frac{2}{5} $
[Solution Description] The number of boys in the class is $30 – 18 = 12$. Thus, the probability of selecting a boy is $\frac{12}{30} = \frac{2}{5}$.

In a scenario where you have a class of students, the probability of selecting a student of a specific gender is determined by dividing the number of students of that gender by the total number of students. For instance, if there are 18 girls in a class of 30 students, the probability of selecting a boy is calculated by subtracting the number of girls from the total and dividing by the total number of students.

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Key Concept: Two Coins Toss

b) 1/2

[Solution Description]

To determine the probability of getting exactly one head when two coins are tossed, let’s first list the possible outcomes: HH, HT, TH, and TT. The favorable outcomes for getting exactly one head are HT and TH.

Number of favorable outcomes = 2 (HT, TH).

Total number of possible outcomes = 4 (HH, HT, TH, TT).

Thus, the theoretical probability is:

$P(\text{exactly one head}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{4} = \frac{1}{2}$

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Key Concept: Multiple Trials

a) $\frac{2}{15}$
[Solution Description]
To find the probability of drawing both green balls, we first find the probability of drawing a green ball on the first draw, which is $\frac{4}{10}$. After removing one green ball, there are now 3 green balls out of 9 total balls left. The probability of drawing a second green ball is then $\frac{3}{9}$. Using the formula for dependent events:

$
P(G_1 \cap G_2) = P(G_1) \times P(G_2|G_1) = \frac{4}{10} \times \frac{3}{9} = \frac{2}{15}
$

Thus, the probability that both balls are green is $\frac{2}{15}$.

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