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Five fair coins are tossed simultaneously. The probability of the events that at least one head comes up is
When five fair coins are tossed simultaneously the probability of getting at least one head is found using the complement rule 1. Total possible outcomes when tossing 5 coins = 2⁵ = 32 2. Only unfavorable outcome is getting all tails which happens in one way: TTTTT 3. Probability of getting all tailRead more
When five fair coins are tossed simultaneously the probability of getting at least one head is found using the complement rule
1. Total possible outcomes when tossing 5 coins = 2⁵ = 32
2. Only unfavorable outcome is getting all tails which happens in one way: TTTTT
3. Probability of getting all tails = 1/32
4. Probability of getting at least one head = 1 – P(all tails) = 1 – 1/32 = 31/32
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If for any two events A and B, P(A) = 4/5 and P(A ∩ B) = 7/10, then P(B/A) is equal to
Given: P(A) = 4/5 P(A ∩ B) = 7/10 Using the conditional probability formula: P(B|A) = P(A ∩ B) / P(A) Substituting the values: P(B|A) = (7/10) ÷ (4/5) = (7/10) × (5/4) = 35/40 = 7/8 So the correct answer is 7/8 Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapteRead more
Given:
P(A) = 4/5
P(A ∩ B) = 7/10
Using the conditional probability formula:
P(B|A) = P(A ∩ B) / P(A)
Substituting the values:
P(B|A) = (7/10) ÷ (4/5)
= (7/10) × (5/4) = 35/40 = 7/8
So the correct answer is 7/8
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Three persons A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is
Given: P(A hits) = 0.4 P(B hits) = 0.3 P(C hits) = 0.2 To find the probability of exactly two hits, we consider the cases where exactly two of A, B, and C hit the target: 1. A and B hit, C misses: P(A ∩ B ∩ C') = (0.4) × (0.3) × (1 - 0.2) = 0.4 × 0.3 × 0.8 = 0.096 2. A and C hit, B misses: P(A ∩ B'Read more
Given:
P(A hits) = 0.4
P(B hits) = 0.3
P(C hits) = 0.2
To find the probability of exactly two hits, we consider the cases where exactly two of A, B, and C hit the target:
1. A and B hit, C misses:
P(A ∩ B ∩ C’) = (0.4) × (0.3) × (1 – 0.2)
= 0.4 × 0.3 × 0.8 = 0.096
2. A and C hit, B misses:
P(A ∩ B’ ∩ C) = (0.4) × (1 – 0.3) × (0.2)
= 0.4 × 0.7 × 0.2 = 0.056
3. B and C hit, A misses:
P(A’ ∩ B ∩ C) = (1 – 0.4) × (0.3) × (0.2)
= 0.6 × 0.3 × 0.2 = 0.036
Total probability of exactly two hits:
P = 0.096 + 0.056 + 0.036 = 0.188
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If the events A and B are independent, then P(A ∩ B) is equal to
For independent events A and B, the probability of both occurring is given by: P(A ∩ B) = P(A) × P(B) This is the fundamental property of independent events in probability. So the correct answer is P(A). P(B) Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-Read more
For independent events A and B, the probability of both occurring is given by:
P(A ∩ B) = P(A) × P(B)
This is the fundamental property of independent events in probability.
So the correct answer is P(A). P(B)
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Two events E and F are independent. If P(E) = 0.3, P(E U F) = 0.5, then P(E/F) – P(F/E) equals
Given: P(E) = 0.3 P(E ∪ F) = 0.5 Since E and F are independent events, we use the formula: P(E ∪ F) = P(E) + P(F) - P(E ∩ F) Given that E and F are independent: P(E ∩ F) = P(E) × P(F) Substitute the values: 0.5 = 0.3 + P(F) - (0.3 × P(F)) 0.5 = 0.3 + P(F) - 0.3P(F) 0.2 = P(F) - 0.3P(F) 0.2 = 0.7P(F)Read more
Given:
P(E) = 0.3
P(E ∪ F) = 0.5
Since E and F are independent events, we use the formula:
P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
Given that E and F are independent:
P(E ∩ F) = P(E) × P(F)
Substitute the values:
0.5 = 0.3 + P(F) – (0.3 × P(F))
0.5 = 0.3 + P(F) – 0.3P(F)
0.2 = P(F) – 0.3P(F)
0.2 = 0.7P(F)
P(F) = 2/7
Now, conditional probability calculations:
P(E | F) = P(E ∩ F) / P(F)
= (0.3 × 2/7) ÷ (2/7)
= 0.3
P(F | E) = P(E ∩ F) / P(E)
= (0.3 × 2/7) ÷ 0.3
= 2/7
Now, computing the needed difference:
P(E | F) – P(F | E) = 0.3 – 2/7
= 3/10 – 2/7
By taking LCM (70):
= (21/70) – (20/70)
= 1/70
Hence, the correct answer is 1/70
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