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Bunuel
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Three coins. two 10-cent coins and one 5-cent coin, are to be flipped 26 Nov 2022, 12:45
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Three coins. two 10-cent coins and one 5-cent coin, are to be flipped simultaneously. For each of the three coins, the probability that the coin will land heads up is 1/2. What is the probability that the total value of the coins that will land heads up is 15 cents?

A. 3/4
B. 1/2
C. 3/8
D. 1/4
E. 1/8
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D
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Three coins. two 10-cent coins and one 5-cent coin, are to be flipped 26 Nov 2022, 13:01
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1/2 x 1/2 x 1/2 = 1/8 probability of one head is 1/2 for three coin 1/8.

Therefore , for 2 heads.
2*(1/8) = 1/4.
Ans D.

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Three coins. two 10-cent coins and one 5-cent coin, are to be flipped 12 Aug 2024, 20:28
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Not a very clear explain explanation
P(H) = 1/2
P(HH)=1/2*1/2
P(HHT)=1/2*1/2*1/2=1/8

We will have to consider the tail part??
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Three coins. two 10-cent coins and one 5-cent coin, are to be flipped 18 Aug 2024, 02:49
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There are three coins: 10-10-5

So we can have total value to be 15 two ways:

(H)*(not H)*(H) + (not H)*(H)*(H)

\(\frac{1}{2}*\frac{1}{2}*\frac{1}{2} + \frac{1}{2}*\frac{1}{2}*\frac{1}{2}\)

We include probability of (not H) because we don't want 3 heads on the tosses.
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Three coins. two 10-cent coins and one 5-cent coin, are to be flipped 24 Apr 2025, 23:08
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We want total 15 cents from:
  • Two 10¢ coins
  • One 5¢ coin
To get 15¢, we must have:
One 10¢ coin = heads
5¢ coin = heads
Other 10¢ coin = tails
How many ways can that happen?
  • The 5¢ coin must be heads → prob = 1/2
  • Exactly one of the two 10¢ coins is heads → two combinations → prob = 2 × (1/2 × 1/2) = 1/2
  • So total probability = (1/2 for 5¢) × (1/2 for 10¢s) = 1/4

✅ Answer: D. 1/4
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Three coins. two 10-cent coins and one 5-cent coin, are to be flipped 24 Apr 2025, 23:15
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Bunuel
Three coins. two 10-cent coins and one 5-cent coin, are to be flipped simultaneously. For each of the three coins, the probability that the coin will land heads up is 1/2. What is the probability that the total value of the coins that will land heads up is 15 cents?

A. 3/4
B. 1/2
C. 3/8
D. 1/4
E. 1/8
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We need heads from one 10-cent coin and one 5-cent coin and the other 10-cent coin should show tails
We can choose one 10-cent coin in 2C1 = 2 ways


Required probability

= (number of ways of choosing the 10-cent coin) × (probability of Head of 10-cent coin) × (probability of Head of 5-cent coin) × (probability of Tail of 10-cent coin)

= 2 × (1/2) × (1/2) × (1/2) = 1/4

Ans D
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