Question

Three coins are tossed together. Find the probability of getting) Exactly one head (ii) At most two heads (iii)At least one head and one tail (iv) No tail

Answer 1

AnSwer: When we'll toss three coins together, then the total number of Outcomes would be 8. (HHH, HHT, THH, HTH, THT, TTH, TTT, HTT)

» To Calculate the probability of any Event ( E ) the formula is Given by :

• Probability = (No. of favourable Outcomes/ Total no. of Outcomes.)

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⌑ The probability of Getting Exactly One Head:

• Number of favourable Outcomes are (THT, TTH, HTT).
• And, Total number of Outcomes are 8.

No. of favourable Outcomes/Total no. of Outcomes = ⅜

∴ The Probability of Getting Exactly One Head is ⅜.

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⌑ The probability of Getting Atmost two Heads:

• Number of favourable Outcomes are (HHH, HHT, THH, HTH).
• And, Total number of Outcomes are 8.

No. of favourable Outcomes/Total no. of Outcomes = ½.

∴ The Probability of Getting Atmost two Heads is ½.

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⌑ The probability of Getting Atleast one head and one tail:

• Number of favourable Outcomes are (HHT, HTH, THH, TTH, THT, HTT).
• And, Total number of Outcomes are 8.

No. of favourable Outcomes/Total no. of Outcomes = ¾

∴ The Probability of Getting Atleast one head and one tail is ¾.

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⌑ The probability of Getting No tail:

• Number of favourable Outcome (HHH)
• And, Total number of Outcomes are 8.

No. of favourable Outcomes/Total no. of Outcomes = ⅛

∴ The Probability of Getting No tail is ⅛.

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Answer 2

$\bf{ \underline{Question : }}$

Three coins are tossed together. Find the probability of getting

1. Exactly one head

2. At most two heads

3. At least one head and one tail

4. No tail

$\bf{ \underline{Answer : }}$

$1 . \sf \: Probability \: of \: getting \: exactly \: one \: head = \frac{3}{8}$

$2. \sf \: Probability \: of \: getting \:at \: most \: two \: heads = \frac{1}{2}$

$3. \sf \: Probability \: of \: getting \:at \: least \: one \: head \: and \: one \: tail = \frac{3}{4}$

$4. \sf \: Probability \: of \: getting \:no \: tail = \frac{1}{8}$

$\bf{ \underline{Solution: }}$

$\sf{ \underline{Formula: }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf P(E) = \frac{No. \: of \: favourable \: outcomes }{Total \: no. \: of \: outcomes}$

Total no. of outcomes = 8 (H,H,H) , (H,H,T) , (H,T,H) , (H,T,T) , (T,H,H) , (T,H,T) , (T,T,H) , (T,T,T)

$1 . \underline{ \sf \: Probability \: of \: getting \: exactly \: one \: head = ?}$

No. of favourable outcomes = 3 (H,T,T) , (T,H,T) , (T,T,H)

$\sf P(E) = \frac{3}{8}$

$2. \underline{\sf \: Probability \: of \: getting \:at \: most \: two \: heads = ?}$

No. of favourable outcomes = 4 (H,H,H) , (H,H,T) , (H,T,H) , (T,H,H)

$\sf P(E) = \frac{4}{8} = \frac{1}{2}$

$3. \underline{\sf \: Probability \: of \: getting \:at \: least \: one \: head \: and \: one \: tail = ?}$

No. of favourable outcomes = 6 (H,H,T) , (H,T,H) , (H,T,T) , (T,H,H) , (T,H,T) , (T,T,H)

$\sf P(E) = \frac{6}{8} = \frac{3}{4}$

$4. \underline{\sf \: Probability \: of \: getting \:no \: tail =?}$

No. of favourable outcomes = 1 (H,H,H)

$\sf P(E) = \frac{1}{8}$