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If 10 coins are tossed 100 times, how many times would you expect 7 coins to fall head upward.
- a)14
- b)12
- c)13
- d)11
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If 10 coins are tossed 100 times, how many times would you expect 7 co...
12 is the answer
Step-by-step explanation:
This is the binomial distribution (p+q)^10 with p =1/2, q = 1/2.
You can have 10 heads, probability of this is p^10 or 1/1024
9 heads 1 tail has pr = 10.p.q^9 = 10/1024
8 heads 2 tails has pr = 10C2.p^2.q^8 = 45/1024
7 heads 3 tails has pr = 10C3.p^7.q^3 = 120/1024
So in 100 trials of this you would expect 7 heads & 3 tails to occur 100 x 120/1024 times
or about 11.7 times. Nearly 12 times.
The nearest whole number is 12.
Most Upvoted Answer
If 10 coins are tossed 100 times, how many times would you expect 7 co...
Expected Value Calculation:
To calculate the expected value, we use the formula:
E(X) = n * p
Where,
n = number of trials
p = probability of success
Here, n = 100 (number of times the coins are tossed)
p = probability of getting heads = 1/2 (since there are two possible outcomes - heads or tails)
So, E(X) = 100 * 1/2 = 50
Now, we need to find the expected number of times 7 coins fall head upward. This is a binomial distribution problem, and we can use the formula:
P(X = k) = nCk * p^k * (1-p)^(n-k)
Where,
n = number of trials
p = probability of success
k = number of successes
Here, n = 10 (number of coins tossed)
p = probability of getting heads = 1/2
k = 7 (number of coins falling head upward)
So, P(X = 7) = 10C7 * (1/2)^7 * (1/2)^3 = 120 * 1/128 * 1/8 = 15/256
Now, to find the expected number of times 7 coins fall head upward, we multiply the probability by the number of trials:
E(X) = n * P(X = k)
E(7 heads out of 10 tosses) = 100 * (15/256) = 1.171875
Rounding this value to the nearest integer, we get 12.
Therefore, the correct answer is option B, 12.
To calculate the expected value, we use the formula:
E(X) = n * p
Where,
n = number of trials
p = probability of success
Here, n = 100 (number of times the coins are tossed)
p = probability of getting heads = 1/2 (since there are two possible outcomes - heads or tails)
So, E(X) = 100 * 1/2 = 50
Now, we need to find the expected number of times 7 coins fall head upward. This is a binomial distribution problem, and we can use the formula:
P(X = k) = nCk * p^k * (1-p)^(n-k)
Where,
n = number of trials
p = probability of success
k = number of successes
Here, n = 10 (number of coins tossed)
p = probability of getting heads = 1/2
k = 7 (number of coins falling head upward)
So, P(X = 7) = 10C7 * (1/2)^7 * (1/2)^3 = 120 * 1/128 * 1/8 = 15/256
Now, to find the expected number of times 7 coins fall head upward, we multiply the probability by the number of trials:
E(X) = n * P(X = k)
E(7 heads out of 10 tosses) = 100 * (15/256) = 1.171875
Rounding this value to the nearest integer, we get 12.
Therefore, the correct answer is option B, 12.
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If 10 coins are tossed 100 times, how many times would you expect 7 coins to fall head upward.a)14b)12c)13d)11Correct answer is option 'B'. Can you explain this answer?
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