Two unbiased coins are tossed simultaneously, then the probability of getting no head is
p/q. Find the value of (p + q)2
Answers
Given :- Two unbiased coins are tossed simultaneously, then the probability of getting no head is p/q. Find the value of (p + q)2 ?
Solution :-
we know that,
- Probability (P) = Favourable number of cases / Total number of cases .
If two unbiased coins are tossed simultaneously, then the total number of possible outcomes may be = HH, HT , TH or TT .
So, here total number of possible cases = 4 .
Now, favourable outcomes for no head means both are tails.
→ No. of favourable outcomes = TT = 1
→ Total number of possible cases = 4
As we know that,
→ Probability (P) = Favourable number of cases / Total number of cases
therefore,
→ Probability of no head = 1/4 .
hence,
→ 1/4 = p/q
- p = 1
- q = 4
so,
→ (p + q)2
→ (1 + 4) * 2
→ 5 * 2
→ 10 (Ans.)
or,
→ (p + q)²
→ (1 + 4)²
→ 5²
→ 25 . (Ans.)
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Given :- Two unbiased coins are tossed simultaneously, then the probability of getting no head is p/q. Find the value of (p + q)2 ?
Solution :-
we know that,
Probability (P) = Favourable number of cases / Total number of cases .
If two unbiased coins are tossed simultaneously, then the total number of possible outcomes may be = HH, HT , TH or TT .
So, here total number of possible cases = 4 .
Now, favourable outcomes for no head means both are tails.
→ No. of favourable outcomes = TT = 1
→ Total number of possible cases = 4
As we know that,
→ Probability (P) = Favourable number of cases / Total number of cases
therefore,
→ Probability of no head = 1/4 .
hence,
→ 1/4 = p/q
p = 1
q = 4
so,
→ (p + q)2
→ (1 + 4) * 2
→ 5 * 2
→ 10 (Ans.)
or,
→ (p + q)²
→ (1 + 4)²
→ 5²
→ 25 . (Ans.)