Question -
Answer -
Since either coin can turn up Head (H) or Tail (T), are thepossible outcomes.
But, now coin is tossed four times so the possible samplespace contains,
S = (HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH,THTH, TTHH,
TTTH, TTHT, THTT, HTTT, TTTT)
As per the condition given the question, a person will winor lose money depending up on the face of the coin so,
(i) For 4 heads = 1 + 1 + 1 + 1 = ₹ 4
So, he wins ₹ 4
(ii) For 3 heads and 1 tail = 1 + 1 + 1 – 1.50
= 3 – 1.50
= ₹ 1.50
So, he will be winning ₹ 1.50
(iii) For 2 heads and 2 tails = 1 + 1 – 1.50 – 1.50
= 2 – 3
= – ₹ 1
So, he will be losing ₹ 1
(iv)For 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50
= 1 – 4.50
= – ₹ 3.50
So, he will be losing Rs. 3.50
(v) For 4 tails = – 1.50 – 1.50 – 1.50 – 1.50
= – ₹ 6
So, he will be losing Rs. 6
Now the sample space of amounts is
S= {4, 1.50, 1.50, 1.50, 1.50, – 1, – 1, – 1, – 1, – 1, – 1,– 3.50, – 3.50, – 3.50, – 3.50, – 6}
Then, n (S) = 16
P (winning ₹ 4) = 1/16
P (winning ₹ 1.50) = 4/16 … [divide both numerator anddenominator by 4]
= ¼
P (winning ₹ 1) = 6/16 … [divide both numerator anddenominator by 2]
= 3/8
P (winning ₹ 3.50) = 4/16 … [divide both numerator anddenominator by 4]
= ¼
P (winning ₹ 6) = 1/16
= 3/8